molympiad.files.wordpress.com · Preface This work blends together classic inequality results with brand new problems, some of which devised only a few days ago. What could be special

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Preface

This work blends together classic inequality results with brand new problems,some of which devised only a few days ago. What could be special about it when somany inequality problem books have already been written? We strongly believe thateven if the topic we plunge into is so general and popular our book is very different.Of course, it is quite easy to say this, so we will give some supporting arguments. Thisbook contains a large variety of problems involving inequalities, most of them difficult,questions that became famous in competitions because of their beauty and difficulty.And, even more importantly, throughout the text we employ our own solutions andpropose a large number of new original problems. There are memorable problemsin this book and memorable solutions as well. This is why this work will clearlyappeal to students who are used to use Cauchy-Schwarz as a verb and want to furtherimprove their algebraic skills and techniques. They will find here tough problems,new results, and even problems that could lead to research. The student who is notas keen in this field will also be exposed to a wide variety of moderate and easyproblems, ideas, techniques, and all the ingredients leading to a good preparationfor mathematical contests. Some of the problems we chose to present are known,but we have included them here with new solutions which show the diversity of ideaspertaining to inequalities. Anyone will find here a challenge to prove his or her skills. Ifwe have not convinced you, then please take a look at the last problems and hopefullyyou will agree with us.

Finally, but not in the end, we would like to extend our deepest appreciationto the proposers of the problems featured in this book and to apologize for notgiving all complete sources, even though we have given our best. Also, we would liketo thank Marian Tetiva, Dung Tran Nam, Constantin Tanasescu, Calin Popa andValentin Vornicu for the beautiful problems they have given us and for the preciouscomments, to Cristian Baba, George Lascu and Calin Popa, for typesetting and forthe many pertinent observations they have provided.

The authors

3

Contents

Preface 3

Chapter 1. Problems 7

Chapter 2. Solutions 25

Glossary 121

Further reading 127

5

CHAPTER 1

Problems

7

8 Problems

1. Prove that the inequality√a2 + (1− b)2 +

√b2 + (1− c)2 +

√c2 + (1− a)2 ≥ 3

√2

2holds for arbitrary real numbers a, b, c.

Komal

2. [ Dinu Serbanescu ] If a, b, c ∈ (0, 1) prove that√

abc +√

(1− a)(1− b)(1− c) < 1.

Junior TST 2002, Romania

3. [ Mircea Lascu ] Let a, b, c be positive real numbers such that abc = 1. Provethat

b + c√a

+c + a√

b+

a + b√c≥√

a +√

b +√

c + 3.

Gazeta Matematica

4. If the equation x4 + ax3 + 2x2 + bx + 1 = 0 has at least one real root, thena2 + b2 ≥ 8.

Tournament of the Towns, 1993

5. Find the maximum value of the expression x3 +y3 +z3−3xyz where x2 +y2 +z2 = 1 and x, y, z are real numbers.

6. Let a, b, c, x, y, z be positive real numbers such that x + y + z = 1. Prove that

ax + by + cz + 2√

(xy + yz + zx)(ab + bc + ca) ≤ a + b + c.

Ukraine, 2001

7. [ Darij Grinberg ] If a, b, c are three positive real numbers, then

a

(b + c)2+

b

(c + a)2+

c

(a + b)2≥ 9

4 (a + b + c).

8. [ Hojoo Lee ] Let a, b, c ≥ 0. Prove that√a4 + a2b2 + b4 +

√b4 + b2c2 + c4 +

√c4 + c2a2 + a4 ≥

≥ a√

2a2 + bc + b√

2b2 + ca + c√

2c2 + ab.

Gazeta Matematica

9. If a, b, c are positive real numbers such that abc = 2, then

a3 + b3 + c3 ≥ a√

b + c + b√

c + a + c√

a + b.

Old and New Inequalities 9

When does equality hold?JBMO 2002 Shortlist

10. [ Ioan Tomescu ] Let x, y, z > 0. Prove thatxyz

(1 + 3x)(x + 8y)(y + 9z)(z + 6)≤ 1

74.

When do we have equality?Gazeta Matematica

11. [ Mihai Piticari, Dan Popescu ] Prove that

5(a2 + b2 + c2) ≤ 6(a3 + b3 + c3) + 1,

for all a, b, c > 0 with a + b + c = 1.

12. [ Mircea Lascu ] Let x1, x2, . . . , xn ∈ R, n ≥ 2 and a > 0 such that x1 +

x2 + . . . + xn = a and x21 + x2

2 + . . . + x2n ≤

a2

n− 1. Prove that xi ∈

[0,

2a

n

], for all

i ∈ {1, 2, . . . , n}.

13. [ Adrian Zahariuc ] Prove that for any a, b, c ∈ (1, 2) the following inequalityholds

b√

a

4b√

c− c√

a+

c√

b

4c√

a− a√

b+

a√

c

4a√

b− b√

c≥ 1.

14. For positive real numbers a, b, c such that abc ≤ 1, prove thata

b+

b

c+

c

a≥ a + b + c.

15. [ Vasile Cırtoaje, Mircea Lascu ] Let a, b, c, x, y, z be positive real numberssuch that a+x ≥ b+y ≥ c+z and a+b+c = x+y+z. Prove that ay+bx ≥ ac+xz.

16. [ Vasile Cırtoaje, Mircea Lascu ] Let a, b, c be positive real numbers so thatabc = 1. Prove that

1 +3

a + b + c≥ 6

ab + ac + bc.


17. Let a, b, c be positive real numbers. Prove that

a3

b2+

b3

c2+

c3

a2≥ a2

b+

b2

c+

c2

a.

JBMO 2002 Shortlist

10 Problems

18. Prove that if n > 3 and x1, x2, . . . , xn > 0 have product 1, then

11 + x1 + x1x2

+1

1 + x2 + x2x3+ · · ·+ 1

1 + xn + xnx1> 1.

Russia, 2004

19. [ Marian Tetiva ] Let x, y, z be positive real numbers satisfying the condition

x2 + y2 + z2 + 2xyz = 1.

Prove thata) xyz ≤ 1

8;

b) x + y + z ≤ 32;

c) xy + xz + yz ≤ 34≤ x2 + y2 + z2;

d) xy + xz + yz ≤ 12

+ 2xyz.

20. [ Marius Olteanu ] Let x1, x2, x3, x4, x5 ∈ R so that x1 +x2 +x3 +x4 +x5 = 0.Prove that

| cos x1|+ | cos x2|+ | cos x3|+ | cos x4|+ | cos x5| ≥ 1.

Gazeta Matematica

21. [ Florina Carlan, Marian Tetiva ] Prove that if x, y, z > 0 satisfy the condition

x + y + z = xyz

then xy + xz + yz ≥ 3 +√

x2 + 1 +√

y2 + 1 +√

z2 + 1.

22. [ Laurentiu Panaitopol ] Prove that

1 + x2

1 + y + z2+

1 + y2

1 + z + x2+

1 + z2

1 + x + y2≥ 2,

for any real numbers x, y, z > −1.JBMO, 2003

23. Let a, b, c > 0 with a + b + c = 1. Show that

a2 + b

b + c+

b2 + c

c + a+

c2 + a

a + b≥ 2.

24. Let a, b, c ≥ 0 such that a4 + b4 + c4 ≤ 2(a2b2 + b2c2 + c2a2). Prove that

a2 + b2 + c2 ≤ 2(ab + bc + ca).

Kvant, 1988


25. Let n ≥ 2 and x1, ..., xn be positive real numbers satisfying1

x1 + 1998+

1x2 + 1998

+ ... +1

xn + 1998=

11998

.

Prove thatn√

x1x2...xn

n− 1≥ 1998.

Vietnam, 1998

26. [ Marian Tetiva ] Consider positive real numbers x, y, z so that

x2 + y2 + z2 = xyz.

Prove the following inequalitiesa) xyz ≥ 27;b) xy + xz + yz ≥ 27;c) x + y + z ≥ 9;d) xy + xz + yz ≥ 2 (x + y + z) + 9.

27. Let x, y, z be positive real numbers with sum 3. Prove that√

x +√

y +√

z ≥ xy + yz + zx.

Russia, 2002

28. [ D. Olteanu ] Let a, b, c be positive real numbers. Prove that

a + b

b + c· a

2a + b + c+

b + c

c + a· b

2b + c + a+

c + a

a + b· c

2c + a + b≥ 3

4.

Gazeta Matematica

29. For any positive real numbers a, b, c show that the following inequality holds

a

b+

b

c+

c

a≥ c + a

c + b+

a + b

a + c+

b + c

b + a.

India, 2002


a3

b2 − bc + c2+

b3

c2 − ac + a2+

c3

a2 − ab + b2≥ 3(ab + bc + ca)

a + b + c.

Proposed for the Balkan Mathematical Olympiad

31. [ Adrian Zahariuc ] Consider the pairwise distinct integers x1, x2, . . . , xn,n ≥ 0. Prove that

x21 + x2

2 + · · ·+ x2n ≥ x1x2 + x2x3 + · · ·+ xnx1 + 2n− 3.

12 Problems

32. [ Murray Klamkin ] Find the maximum value of the expression x21x2 +x2

2x3 +· · ·+ x2

n−1xn + x2nx1 when x1, x2, . . . , xn−1, xn ≥ 0 add up to 1 and n > 2.

Crux Mathematicorum

33. Find the maximum value of the constant c such that for anyx1, x2, . . . , xn, · · · > 0 for which xk+1 ≥ x1 + x2 + · · ·+ xk for any k, the inequality

√x1 +

√x2 + · · ·+

√xn ≤ c

√x1 + x2 + · · ·+ xn

also holds for any n.IMO Shortlist, 1986

34. Given are positive real numbers a, b, c and x, y, z, for which a + x = b + y =c + z = 1. Prove that

(abc + xyz)( 1

ay+

1bz

+1cx

)≥ 3.

Russia, 2002

35. [ Viorel Vajaitu, Alexandru Zaharescu ] Let a, b, c be positive real numbers.Prove that

ab

a + b + 2c+

bc

b + c + 2a+

ca

c + a + 2b≤ 1

4(a + b + c).

Gazeta Matematica

36. Find the maximum value of the expression

a3(b + c + d) + b3(c + d + a) + c3(d + a + b) + d3(a + b + c)

where a, b, c, d are real numbers whose sum of squares is 1.

37. [ Walther Janous ] Let x, y, z be positive real numbers. Prove thatx

x +√

(x + y)(x + z)+

y

y +√

(y + z)(y + x)+

z

z +√

(z + x)(z + y)≤ 1.

Crux Mathematicorum

38. Suppose that a1 < a2 < ... < an are real numbers for some integer n ≥ 2.Prove that

a1a42 + a2a

43 + ... + ana4

1 ≥ a2a41 + a3a

42 + ... + a1a

4n.

Iran, 1999

39. [ Mircea Lascu ] Let a, b, c be positive real numbers. Prove that

b + c

a+

c + a

b+

a + b

c≥ 4

(a

b + c+

b

c + a+

c

a + b

).


40. Let a1, a2, . . . , an > 1 be positive integers. Prove that at least one of thenumbers a1

√a2, a2

√a3, . . . , an−1

√an, an

√a1 is less than or equal to 3

√3.

Adapted after a well-known problem

41. [ Mircea Lascu, Marian Tetiva ] Let x, y, z be positive real numbers whichsatisfy the condition

xy + xz + yz + 2xyz = 1.

Prove that the following inequalities hold

a) xyz ≤ 18;

b) x + y + z ≥ 32;

c)1x

+1y

+1z≥ 4 (x + y + z);

d)1x

+1y

+1z− 4 (x + y + z) ≥ (2z − 1)2

(z (2z + 1)), where z = max {x, y, z}.

42. [ Manlio Marangelli ] Prove that for any positive real numbers x, y, z,

3(x2y + y2z + z2x)(xy2 + yz2 + zx2) ≥ xyz(x + y + z)3.

43. [ Gabriel Dospinescu ] Prove that if a, b, c are real numbers such thatmax{a, b, c} −min{a, b, c} ≤ 1, then

1 + a3 + b3 + c3 + 6abc ≥ 3a2b + 3b2c + 3c2a

44. [ Gabriel Dospinescu ] Prove that for any positive real numbers a, b, c we have

27 +(

2 +a2

bc

)(2 +

b2

ca

)(2 +

c2

ab

)≥ 6(a + b + c)

(1a

+1b

+1c

).

45. Let a0 =12

and ak+1 = ak +a2

k

n. Prove that 1− 1

n< an < 1.

TST Singapore

46. [ Calin Popa ] Let a, b, c be positive real numbers, with a, b, c ∈ (0, 1) suchthat ab + bc + ca = 1. Prove that

a

1− a2+

b

1− b2+

c

1− c2≥ 3

4

(1− a2

a+

1− b2

b+

1− c2

c

).

47. [ Titu Andreescu, Gabriel Dospinescu ] Let x, y, z ≤ 1 and x + y + z = 1.Prove that

11 + x2

+1

1 + y2+

11 + z2

≤ 2710

.

14 Problems

48. [ Gabriel Dospinescu ] Prove that if√

x +√

y +√

z = 1, then

(1− x)2(1− y)2(1− z)2 ≥ 215xyz(x + y)(y + z)(z + x)

.

49. Let x, y, z be positive real numbers such that xyz = x+ y + z +2. Prove that

(1) xy + yz + zx ≥ 2(x + y + z);

(2)√

x +√

y +√

z ≤ 32√

xyz.

50. Prove that if x, y, z are real numbers such that x2 + y2 + z2 = 2, then

x + y + z ≤ xyz + 2.

IMO Shortlist, 1987

51. [ Titu Andreescu, Gabriel Dospinescu ] Prove that for any x1, x2, . . . , xn ∈(0, 1) and for any permutation σ of the set {1, 2, . . . , n}, we have the inequality

n∑i=1

11− xi

≥

1 +

n∑i=1

xi

n

·(

n∑i=1

11− xi · xσ(i)

).

52. Let x1, x2, . . . , xn be positive real numbers such thatn∑

i=1

11 + xi

= 1. Prove

thatn∑

i=1

√xi ≥ (n− 1)

n∑i=1

1√

xi.

Vojtech Jarnik

53. [ Titu Andreescu ] Let n > 3 and a1, a2, . . . , an be real numbers such thata1 +a2 + . . .+an ≥ n and a2

1 +a22 + . . .+a2

n ≥ n2. Prove that max{a1, a2, . . . , an} ≥ 2.

USAMO, 1999

54. [ Vasile Cırtoaje ] If a, b, c, d are positive real numbers, then

a− b

b + c+

b− c

c + d+

c− d

d + a+

d− a

a + b≥ 0.

55. If x and y are positive real numbers, show that xy + yx > 1.France, 1996


56. Prove that if a, b, c > 0 have product 1, then

(a + b)(b + c)(c + a) ≥ 4(a + b + c− 1).

MOSP, 2001

57. Prove that for any a, b, c > 0,

(a2 + b2 + c2)(a + b− c)(b + c− a)(c + a− b) ≤ abc(ab + bc + ca).

58. [ D.P.Mavlo ] Let a, b, c > 0. Prove that

3 + a + b + c +1a

+1b

+1c

+a

b+

b

c+

c

a≥ 3

(a + 1)(b + 1)(c + 1)1 + abc

.

Kvant, 1988

59. [ Gabriel Dospinescu ] Prove that for any positive real numbers x1, x2, . . . , xn

with product 1 we have the inequality

nn ·n∏

i=1

(xni + 1) ≥

(n∑

i=1

xi +n∑

i=1

1xi

)n

.

60. Let a, b, c, d > 0 such that a + b + c = 1. Prove that

a3 + b3 + c3 + abcd ≥ min{

14,19

+d

27

}.

Kvant, 1993

61. Prove that for any real numbers a, b, c we have the inequality∑(1 + a2)2(1 + b2)2(a− c)2(b− c)2 ≥ (1 + a2)(1 + b2)(1 + c2)(a− b)2(b− c)2(c− a)2.

AMM

62. [ Titu Andreescu, Mircea Lascu ] Let α, x, y, z be positive real numbers suchthat xyz = 1 and α ≥ 1. Prove that

xα

y + z+

yα

z + x+

zα

x + y≥ 3

2.

63. Prove that for any real numbers x1, . . . , xn, y1, . . . , yn such that x21+· · ·+x2

n =y21 + · · ·+ y2

n = 1,

(x1y2 − x2y1)2 ≤ 2

(1−

n∑k=1

xkyk

).

Korea, 2001

16 Problems

64. [ Laurentiu Panaitopol ] Let a1, a2, . . . , an be pairwise distinct positive inte-gers. Prove that

a21 + a2

2 + · · ·+ a2n ≥

2n + 13

(a1 + a2 + · · ·+ an).

TST Romania

65. [ Calin Popa ] Let a, b, c be positive real numbers such that a + b + c = 1.Prove that

b√

c

a(√

3c +√

ab)+

c√

a

b(√

3a +√

bc)+

a√

b

c(√

3b +√

ca)≥ 3

√3

4.

66. [ Titu Andreescu, Gabriel Dospinescu ] Let a, b, c, d be real numbers such that(1 + a2)(1 + b2)(1 + c2)(1 + d2) = 16. Prove that

−3 ≤ ab + bc + cd + da + ac + bd− abcd ≤ 5.

67. Prove that

(a2 + 2)(b2 + 2)(c2 + 2) ≥ 9(ab + bc + ca)

for any positive real numbers a, b, c.APMO, 2004

68. [ Vasile Cırtoaje ] Prove that if 0 < x ≤ y ≤ z and x + y + z = xyz + 2, thena) (1− xy)(1− yz)(1− xz) ≥ 0;

b) x2y ≤ 1, x3y2 ≤ 3227

.

69. [ Titu Andreescu ] Let a, b, c be positive real numbers such that a+b+c ≥ abc.

Prove that at least two of the inequalities

2a

+3b

+6c≥ 6,

2b

+3c

+6a≥ 6,

2c

+3a

+6b≥ 6,

are true.TST 2001, USA

70. [ Gabriel Dospinescu, Marian Tetiva ] Let x, y, z > 0 such that

x + y + z = xyz.

Prove that

(x− 1) (y − 1) (z − 1) ≤ 6√

3− 10.


71. [ Marian Tetiva ] Prove that for any positive real numbers a, b, c,∣∣∣∣a3 − b3

a + b+

b3 − c3

b + c+

c3 − a3

c + a

∣∣∣∣ ≤ (a− b)2 + (b− c)2 + (c− a)2

4.

Moldova TST, 2004

72. [ Titu Andreescu ] Let a, b, c be positive real numbers. Prove that

(a5 − a2 + 3)(b5 − b2 + 3)(c5 − c2 + 3) ≥ (a + b + c)3.

USAMO, 2004

73. [ Gabriel Dospinescu ] Let n > 2 and x1, x2, . . . , xn > 0 such that(n∑

k=1

xk

)(n∑

k=1

1xk

)= n2 + 1.

Prove that (n∑

k=1

x2k

)·

(n∑

k=1

1x2

k

)> n2 + 4 +

2n(n− 1)

.

74. [ Gabriel Dospinescu, Mircea Lascu, Marian Tetiva ] Prove that for any po-sitive real numbers a, b, c,

a2 + b2 + c2 + 2abc + 3 ≥ (1 + a)(1 + b)(1 + c).

75. [ Titu Andreescu, Zuming Feng ] Let a, b, c be positive real numbers. Provethat

(2a + b + c)2

2a2 + (b + c)2+

(2b + a + c)2

2b2 + (a + c)2+

(2c + a + b)2

2c2 + (a + b)2≤ 8.

USAMO, 2003

76. Prove that for any positive real numbers x, y and any positive integers m,n,

(n−1)(m−1)(xm+n+ym+n)+(m+n−1)(xmyn+xnym) ≥ mn(xm+n−1y+ym+n−1x).

Austrian-Polish Competition, 1995

77. Let a, b, c, d, e be positive real numbers such that abcde = 1. Prove that

a + abc

1 + ab + abcd+

b + bcd

1 + bc + bcde+

c + cde

1 + cd + cdea+

d + dea

1 + de + deab+

e + eab

1 + ea + eabc≥ 10

3.

Crux Mathematicorum

18 Problems

78. [ Titu Andreescu ] Prove that for any a, b, c, ∈(0,

π

2

)the following inequality

holdssin a · sin(a− b) · sin(a− c)

sin(b + c)+

sin b · sin(b− c) · sin(b− a)sin(c + a)

+sin c · sin(c− a) · sin(c− b)

sin(a + b)≥ 0.

TST 2003, USA

79. Prove that if a, b, c are positive real numbers then,√a4 + b4 + c4 +

√a2b2 + b2c2 + c2a2 ≥

√a3b + b3c + c3a +

√ab3 + bc3 + ca3.

KMO Summer Program Test, 2001

80. [ Gabriel Dospinescu, Mircea Lascu ] For a given n > 2 find the smallestconstant kn with the property: if a1, . . . , an > 0 have product 1, then

a1a2

(a21 + a2)(a2

2 + a1)+

a2a3

(a22 + a3)(a2

3 + a2)+ · · ·+ ana1

(a2n + a1)(a2

1 + an)≤ kn.

81. [ Vasile Cırtoaje ] For any real numbers a, b, c, x, y, z prove that the inequalityholds

ax + by + cz +√

(a2 + b2 + c2)(x2 + y2 + z2) ≥ 23(a + b + c)(x + y + z).

Kvant, 1989

82. [ Vasile Cırtoaje ] Prove that the sides a, b, c of a triangle satisfy the inequality

3(

a

b+

b

c+

c

a− 1)≥ 2

(b

a+

c

b+

a

c

).

83. [ Walther Janous ] Let n > 2 and let x1, x2, . . . , xn > 0 add up to 1. Provethat

n∏i=1

(1 +

1xi

)≥

n∏i=1

(n− xi

1− xi

).

Crux Mathematicorum

84. [ Vasile Cırtoaje, Gheorghe Eckstein ] Consider positive real numbersx1, x2, ..., xn such that x1x2...xn = 1. Prove that

1n− 1 + x1

+1

n− 1 + x2+ ... +

1n− 1 + xn

≤ 1.

TST 1999, Romania

85. [ Titu Andreescu ] Prove that for any nonnegative real numbers a, b, c suchthat a2 + b2 + c2 + abc = 4 we have 0 ≤ ab + bc + ca− abc ≤ 2.

USAMO, 2001


86. [ Titu Andreescu ] Prove that for any positive real numbers a, b, c the followinginequality holds

a + b + c

3− 3√

abc ≤ max{(√

a−√

b)2, (√

b−√

c)2, (√

c−√

a)2}.

TST 2000, USA

87. [ Kiran Kedlaya ] Let a,b,c be positive real numbers. Prove that

a +√

ab + 3√

abc

3≤ 3

√a · a + b

2· a + b + c

3.

88. Find the greatest constant k such that for any positive integer n which is nota square, |(1 +

√n) sin(π

√n)| > k.

Vietnamese IMO Training Camp, 1995

89. [ Dung Tran Nam ] Let x, y, z > 0 such that (x + y + z)3 = 32xyz. Find the

minimum and maximum ofx4 + y4 + z4

(x + y + z)4.

Vietnam, 2004

90. [ George Tsintifas ] Prove that for any a, b, c, d > 0,

(a + b)3(b + c)3(c + d)3(d + a)3 ≥ 16a2b2c2d2(a + b + c + d)4.

Crux Mathematicorum

91. [ Titu Andreescu, Gabriel Dospinescu ] Find the maximum value of the ex-pression

(ab)n

1− ab+

(bc)n

1− bc+

(ca)n

1− ca

where a, b, c are nonnegative real numbers which add up to 1 and n is some positiveinteger.


1a(1 + b)

+1

b(1 + c)+

1c(1 + a)

≥ 33√

abc(1 + 3√

abc).

93. [ Dung Tran Nam ] Prove that for any real numbers a, b, c such that a2 + b2 +c2 = 9,

2(a + b + c)− abc ≤ 10.

Vietnam, 2002

20 Problems

94. [ Vasile Cırtoaje ] Let a, b, c be positive real numbers. Prove that(a +

1b− 1)(

b +1c− 1)

+(

b +1c− 1)(

c +1a− 1)

+(

c +1a− 1)(

a +1b− 1)≥ 3.

95. [ Gabriel Dospinescu ] Let n be an integer greater than 2. Find the greatestreal number mn and the least real number Mn such that for any positive real numbersx1, x2, . . . , xn (with xn = x0, xn+1 = x1),

mn ≤n∑

i=1

xi

xi−1 + 2(n− 1)xi + xi+1≤ Mn.

96. [ Vasile Cırtoaje ] If x, y, z are positive real numbers, then

1x2 + xy + y2

+1

y2 + yz + z2+

1z2 + zx + z2

≥ 9(x + y + z)2

.

Gazeta Matematica

97. [ Vasile Cırtoaje ] For any a, b, c, d > 0 prove that

2(a3 + 1)(b3 + 1)(c3 + 1)(d3 + 1) ≥ (1 + abcd)(1 + a2)(1 + b2)(1 + c2)(1 + d2).

Gazeta Matematica

98. Prove that for any real numbers a, b, c,

(a + b)4 + (b + c)4 + (c + a)4 ≥ 47(a4 + b4 + c4).

Vietnam TST, 1996

99. Prove that if a, b, c are positive real numbers such that abc = 1, then1

1 + a + b+

11 + b + c

+1

1 + c + a≤ 1

2 + a+

12 + b

+1

2 + c.

Bulgaria, 1997

100. [ Dung Tran Nam ] Find the minimum value of the expression1a

+2b

+3c

where a, b, c are positive real numbers such that 21ab + 2bc + 8ca ≤ 12.Vietnam, 2001

101. [ Titu Andreescu, Gabriel Dospinescu ] Prove that for any x, y, z, a, b, c > 0such that xy + yz + zx = 3,

a

b + c(y + z) +

b

c + a(z + x) +

c

a + b(x + y) ≥ 3.



(b + c− a)2

(b + c)2 + a2+

(c + a− b)2

(c + a)2 + b2+

(a + b− c)2

(a + b)2 + c2≥ 3

5.

Japan, 1997

103. [ Vasile Cırtoaje, Gabriel Dospinescu ] Prove that if a1, a2, . . . , an ≥ 0 then

an1 + an

2 + · · ·+ ann − na1a2 . . . an ≥ (n− 1)

(a1 + a2 + · · ·+ an−1

n− 1− an

)n

where an is the least among the numbers a1, a2, . . . , an.

104. [ Turkevici ] Prove that for all positive real numbers x, y, z, t,

x4 + y4 + z4 + t4 + 2xyzt ≥ x2y2 + y2z2 + z2t2 + t2x2 + x2z2 + y2t2.

Kvant

105. Prove that for any real numbers a1, a2, . . . , an the following inequality holds(n∑

i=1

ai

)2

≤n∑

i,j=1

ij

i + j − 1aiaj .

106. Prove that if a1, a2, . . . , an, b1, . . . , bn are real numbers between 1001 and2002, inclusively, such that a2

1 + a22 + · · ·+ a2

n = b21 + b2

2 + · · ·+ b2n, then we have the

inequalitya31

b1+

a32

b2+ · · ·+ a3

n

bn≤ 17

10(a21 + a2

2 + · · ·+ a2n

).

TST Singapore

107. [ Titu Andreescu, Gabriel Dospinescu ] Prove that if a, b, c are positive realnumbers which add up to 1, then

(a2 + b2)(b2 + c2)(c2 + a2) ≥ 8(a2b2 + b2c2 + c2a2)2.

108. [ Vasile Cırtoaje ] If a, b, c, d are positive real numbers such that abcd = 1,then

1(1 + a)2

+1

(1 + b)2+

1(1 + c)2

+1

(1 + d)2≥ 1.

Gazeta Matematica

109. [ Vasile Cırtoaje ] Let a, b, c be positive real numbers. Prove that

a2

b2 + c2+

b2

c2 + a2+

c2

a2 + b2≥ a

b + c+

b

c + a+

c

a + b.

Gazeta Matematica

22 Problems

110. [ Gabriel Dospinescu ] Let a1, a2, . . . , an be real numbers and let S be anon-empty subset of {1, 2, . . . , n}. Prove that(∑

i∈S

ai

)2

≤∑

1≤i≤j≤n

(ai + . . . + aj)2.

TST 2004, Romania

111. [ Dung Tran Nam ] Let x1, x2 . . . , x2004 be real numbers in the interval [−1, 1]such that x3

1 +x32 + . . .+x3

2004 = 0. Find the maximal value of the x1 +x2 + · · ·+x2004.

112. [ Gabriel Dospinescu, Calin Popa ] Prove that if n ≥ 2 and a1, a2, . . . , an

are real numbers with product 1, then

a21 + a2

2 + · · ·+ a2n − n ≥ 2n

n− 1· n√

n− 1(a1 + a2 + · · ·+ an − n).

113. [ Vasile Cırtoaje ] If a, b, c are positive real numbers, then√2a

a + b+

√2b

b + c+

√2c

c + a≤ 3.

Gazeta Matematica

114. Prove the following inequality for positive real numbers x, y, z

(xy + yz + zx)(

1(x + y)2

+1

(y + z)2+

1(z + x)2

)≥ 9

4

Iran, 1996

115. Prove that for any x, y in the interval [0, 1],√1 + x2 +

√1 + y2 +

√(1− x)2 + (1− y)2 ≥ (1 +

√5)(1− xy).

116. [ Suranyi ] Prove that for any positive real numbers a1, a2, . . . , an the fol-lowing inequality holds

(n−1)(an1 +an

2 +· · ·+ann)+na1a2 . . . an ≥ (a1+a2+· · ·+an)(an−1

1 +an−12 +· · ·+an−1

n ).

Miklos Schweitzer Competition

117. Prove that for any x1, x2, . . . , xn > 0 with product 1,∑1≤i<j≤n

(xi − xj)2 ≥n∑

i=1

x2i − n.

A generalization of Turkevici’s inequality


118. [ Gabriel Dospinescu ] Find the minimum value of the expressionn∑

i=1

√a1a2 . . . an

1− (n− 1)ai

where a1, a2, . . . , an <1

n− 1add up to 1 and n > 2 is an integer.

119. [ Vasile Cırtoaje ] Let a1, a2, . . . , an < 1 be nonnegative real numbers suchthat

a =

√a21 + a2

2 + . . . + a2n

n≥√

33

.

Prove thata1

1− a21

+a2

1− a22

+ . . . +an

1− a2n

≥ na

1− a2.

120. [ Vasile Cırtoaje, Mircea Lascu ] Let a, b, c, x, y, z be positive real numberssuch that

(a + b + c)(x + y + z) = (a2 + b2 + c2)(x2 + y2 + z2) = 4.

Prove thatabcxyz <

136

.

121. [ Gabriel Dospinescu ] For a given n > 2, find the minimal value of theconstant kn, such that if x1, x2, . . . , xn > 0 have product 1, then

1√1 + knx1

+1√

1 + knx2

+ · · ·+ 1√1 + knxn

≤ n− 1.

Mathlinks Contest

122. [ Vasile Cırtoaje, Gabriel Dospinescu ] For a given n > 2, find the maximalvalue of the constant kn such that for any x1, x2, . . . , xn > 0 for which x2

1 +x22 + · · ·+

x2n = 1 we have the inequality

(1− x1)(1− x2) . . . (1− xn) ≥ knx1x2 . . . xn.

CHAPTER 2

Solutions

25

26 Solutions

1. Prove that the inequality√a2 + (1− b)2 +

√b2 + (1− c)2 +

√c2 + (1− a)2 ≥ 3

√2

2holds for arbitrary real numbers a, b, c.

Komal

First solution:Applying Minkowsky’s Inequality to the left-hand side we have√

a2 + (1− b)2+√

b2 + (1− c)2+√

c2 + (1− a)2 ≥√

(a + b + c)2 + (3− a− b− c)2.

Denoting a + b + c = x we get

(a + b + c)2 + (3− a− b− c)2 = 2(

x− 32

)2

+92≥ 9

2,

and the conclusion follows.

Second solution:We have the inequalities√

a2 + (1− b)2 +√

b2 + (1− c)2 +√

c2 + (1− a)2 ≥

≥ |a|+ |1− b|√2

+|b|+ |1− c|√

2+|c|+ |1− a|√

2

and because |x|+ |1− x| ≥ 1 for all real numbers x the last quantity is at least3√

22

.

2. [ Dinu Serbanescu ] If a, b, c ∈ (0, 1) prove that√

abc +√

(1− a)(1− b)(1− c) < 1.


First solution:Observe that x

12 < x

13 for x ∈ (0, 1). Thus

√abc <

3√

abc,

and √(1− a)(1− b)(1− c) < 3

√(1− a)(1− b)(1− c).

By the AM-GM Inequality,√

abc <3√

abc ≤ a + b + c

3,

and√(1− a)(1− b)(1− c) < 3

√(1− a)(1− b)(1− c) ≤ (1− a) + (1− b) + (1− c)

3.


Summing up, we obtain√

abc +√

(1− a)(1− b)(1− c) <a + b + c + 1− a + 1− b + 1− c

3= 1,

as desired.

Second solution:We have

√abc +

√(1− a)(1− b)(1− c) <

√b ·√

c +√

1− b ·√

1− c ≤ 1,

by the Cauchy-Schwarz Inequality.

Third solution:Let a = sin2 x, b = sin2 y, c = sin2 z, where x, y, z ∈

(0,

π

2

). The inequality

becomessinx · sin y · sin z + cos x · cos y · cos z < 1

and it follows from the inequalities

sinx · sin y · sin z + cos x · cos y · cos z < sinx · sin y + cos x · cos y = cos(x− y) ≤ 1

3. [ Mircea Lascu ] Let a, b, c be positive real numbers such that abc = 1. Provethat

b + c√a

+c + a√

b+

a + b√c≥√

a +√

b +√

c + 3.

Gazeta Matematica

Solution:From the AM-GM Inequality, we have

b + c√a

+c + a√

b+

a + b√c≥ 2

(√bc

a+√

ca

b+

√ab

c

)=

=

(√bc

a+√

ca

b

)+

(√ca

b+

√ab

c

)+

(√ab

c+

√bc

a

)≥

≥ 2(√

a +√

b +√

c) ≥√

a +√

b +√

c + 3 6√

abc =√

a +√

b +√

c + 3.

4. If the equation x4 + ax3 + 2x2 + bx + 1 = 0 has at least one real root, thena2 + b2 ≥ 8.

Tournament of the Towns, 1993

28 Solutions

Solution:Let x be the real root of the equation. Using the Cauchy-Schwarz Inequality

we infer that

a2 + b2 ≥(x4 + 2x2 + 1

)2x2 + x6

≥ 8,

because the last inequality is equivalent to (x2 − 1)4 ≥ 0.

5. Find the maximum value of the expression x3 +y3 +z3−3xyz where x2 +y2 +z2 = 1 and x, y, z are real numbers.

Solution:Let t = xy + yz + zx. Let us observe that

(x3 + y3 + z3 − 3xyz)2 = (x + y + z)2(1− xy − yz − zx)2 = (1 + 2t)(1− t)2.

We also have −12≤ t ≤ 1. Thus, we must find the maximum value of the expression

(1+2t)(1− t)2, where −12≤ t ≤ 1. In this case we clearly have (1+2t)(t−1)2 ≤ 1 ⇔

t2(3−2t) ≥ 0 and thus |x3+y3+z3−3xyz| ≤ 1. We have equality for x = 1, y = z = 0and thus the maximum value is 1.

6. Let a, b, c, x, y, z be positive real numbers such that x + y + z = 1. Prove that

ax + by + cz + 2√

(xy + yz + zx)(ab + bc + ca) ≤ a + b + c.

Ukraine, 2001

First solution:We will use the Cauchy-Schwarz Inequality twice. First, we can write ax+by+

cz ≤√

a2 + b2 + c2 ·√

x2 + y2 + z2 and then we apply again the Cauchy-SchwarzInequality to obtain:

ax + by + cz + 2√

(xy + yz + zx)(ab + bc + ca) ≤

≤√∑

a2 ·√∑

x2 +√

2∑

ab ·√

2∑

xy ≤

≤√∑

x2 + 2∑

xy ·√∑

a2 + 2∑

ab =∑

a.

Second solution:The inequality being homogeneous in a, b, c we can assume that a+ b+ c = 1. We

apply this time the AM-GM Inequality and we find that

ax+by+cz+2√

(xy + yz + zx)(ab + bc + ca) ≤ ax+by+cz+xy+yz+zx+ab+bc+ca.

Consequently,

xy + yz + zx + ab + bc + ca =1− x2 − y2 − z2

2+

1− a2 − b2 − c2

2≤ 1− ax− by− cz,


the last one being equivalent to (x− a)2 + (y − b)2 + (z − c)2 ≥ 0.

7. [ Darij Grinberg ] If a, b, c are three positive real numbers, then

a

(b + c)2+

b

(c + a)2+

c

(a + b)2≥ 9

4 (a + b + c).

First solution:We rewrite the inequality as

(a + b + c)

(a

(b + c)2+

b

(c + a)2+

c

(a + b)2

)≥ 9

4.

Applying the Cauchy-Schwarz Inequality we get

(a + b + c)

(a

(b + c)2+

b

(c + a)2+

c

(a + b)2

)≥(

a

b + c+

b

c + a+

c

a + b

)2

.

It remains to prove thata

b + c+

b

c + a+

c

a + b≥ 3

2which is well-known.

Second solution:Without loss of generality we may assume that a + b + c = 1. Now, consider

the function f : (0, 1) → (0,∞), f(x) =x

(1− x)2. A short computation of derivatives

shows that f is convex. Thus, we may apply Jensen’s Inequality and the conclusionfollows.

8. [ Hojoo Lee ] Let a, b, c ≥ 0. Prove that√a4 + a2b2 + b4 +

√b4 + b2c2 + c4 +

√c4 + c2a2 + a4 ≥

≥ a√

2a2 + bc + b√

2b2 + ca + c√

2c2 + ab.

Gazeta Matematica

Solution:We start from (a2−b2)2 ≥ 0. We rewrite it as 4a4+4a2b2+4b4 ≥ 3a4+6a2b2+3b4.

It follows that√

a4 + a2b2 + b2 ≥√

32

(a2 + b2).Using this observation, we find that(∑√

a4 + a2b2 + b4)2

≥ 3(∑

a2)2

.

But using the Cauchy-Schwarz Inequality we obtain(∑a√

2a2 + bc)2

≤(∑

a2)(∑(

2a2 + bc))≤ 3

(∑a2)2

and the inequality is proved.

30 Solutions

9. If a, b, c are positive real numbers such that abc = 2, then

a3 + b3 + c3 ≥ a√

b + c + b√

c + a + c√

a + b.

When does equality hold?JBMO 2002 Shortlist

First solution:Applying the Cauchy-Schwarz Inequality gives

3(a2 + b2 + c2) ≥ 3(a + b + c)2 (1)

and(a2 + b2 + c2)2 ≤ (a + b + c)(a3 + b3 + c3). (2)

These two inequalities combined yield

a3 + b3 + c3 ≥ (a2 + b2 + c2)(a + b + c)3

=(a2 + b2 + c2)[(b + c) + (a + c) + (a + b)]

6

≥ (a√

b + c + b√

a + c + c√

a + b)2

6(3)

Using the AM-GM Inequality we obtain

a√

b + c + b√

a + c + c√

a + b ≥ 3 3

√abc(√

(a + b)(b + c)(c + a))

≥ 33√

abc√

8abc = 3 3√

8 = 6.

Thus

(a√

b + c + b√

a + c + c√

a + b)2 ≥ 6(a√

b + c + b√

a + c + c√

a + b). (4)

The desired inequality follows from (3) and (4).

Second solution:We have

a√

b + c + b√

c + a + c√

a + b ≤√

2(a2 + b2 + c2)(a + b + c).

Using Chebyshev Inequality, we infer that√2(a2 + b2 + c2)(a + b + c) ≤

√6(a3 + b3 + c3)

and so it is enough to prove that a3 + b3 + c3 ≥ 3abc, which is true by the AM-GMInequality. We have equality if a = b = c = 3

√2.

10. [ Ioan Tomescu ] Let x, y, z > 0. Prove that

xyz

(1 + 3x)(x + 8y)(y + 9z)(z + 6)≤ 1

74.


When do we have equality?Gazeta Matematica

Solution:First, we write the inequality in the following form

(1 + 3x)(

1 +8y

x

)(1 +

9z

y

)(1 +

6z

)≥ 74.

But this follows immediately from Huygens Inequality. We have equality for

x = 2, y =32, z = 1.

11. [ Mihai Piticari, Dan Popescu ] Prove that

5(a2 + b2 + c2) ≤ 6(a3 + b3 + c3) + 1,

for all a, b, c > 0 with a + b + c = 1.

Solution:Because a + b + c = 1, we have a3 + b3 + c3 = 3abc + a2 + b2 + c2 − ab− bc− ca.

The inequality becomes

5(a2 + b2 + c2) ≤ 18abc + 6(a2 + b2 + c2)− 6(ab + bc + ca) + 1 ⇔

⇔ 18abc + 1− 2(ab + bc + ca) + 1 ≥ 6(ab + bc + ca) ⇔

⇔ 8(ab + bc + ca) ≤ 2 + 18abc ⇔ 4(ab + bc + ca) ≤ 1 + 9abc ⇔

⇔ (1− 2a)(1− 2b)(1− 2c) ≤ abc ⇔

⇔ (b + c− a)(c + a− b)(a + b− c) ≤ abc,

which is equivalent to Schur’s Inequality.

12. [ Mircea Lascu ] Let x1, x2, . . . , xn ∈ R, n ≥ 2 and a > 0 such that x1 +

x2 + . . . + xn = a and x21 + x2

2 + . . . + x2n ≤

a2

n− 1. Prove that xi ∈

[0,

2a

n

], for all

i ∈ {1, 2, . . . , n}.

Solution:Using the Cauchy-Schwarz Inequality, we get

(a− x1)2 ≤ (n− 1)(x2

2 + x23 + · · ·+ x2

n

)≤ (n− 1)

(a2

n− 1− x2

1

).

Thus,

a2 − 2ax1 + x21 ≤ a2 − (n− 1)x2

1 ⇔ x1

(x1 −

2a

n

)≤ 0


32 Solutions

13. [ Adrian Zahariuc ] Prove that for any a, b, c ∈ (1, 2) the following inequalityholds

b√

a

4b√

c− c√

a+

c√

b

4c√

a− a√

b+

a√

c

4a√

b− b√

c≥ 1.

Solution:The fact that a, b, c ∈ (1, 2) makes all denominators positive. Then

b√

a

4b√

c− c√

a≥ a

a + b + c⇔ b(a + b + c) ≥

√a(4b

√c− c

√a) ⇔

⇔ (a + b)(b + c) ≥ 4b√

ac,

the last one coming from a + b ≥ 2√

ab and b + c ≥ 2√

bc. Writing the other twoinequalities and adding them up give the desired result.

14. For positive real numbers a, b, c such that abc ≤ 1, prove that

a

b+

b

c+

c

a≥ a + b + c.

First solution:If ab + bc + ca ≤ a + b + c, then the Cauchy-Schwarz Inequality solves the

problem:

a

b+

b

c+

c

a=

a2c + b2a + c2b

abc≥

(a + b + c)2

1a

+1b

+1c

abc≥ a + b + c.

Otherwise, the same inequality gives

a

b+

b

c+

c

a=

a2c + b2a + c2b

abc≥

(ab + bc + ca)2

a + b + cabc

≥ a + b + c

(here we have used the fact that abc ≤ 1).

Second solution:Replacing a, b, c by ta, tb, tc with t =

13√

abcpreserves the value of the quantity in

the left-hand side of the inequality and increases the value of the right-hand side andmakes at · bt · ct = abct3 = 1. Hence we may assume without loss of generality thatabc = 1. Then there exist positive real numbers x, y, z such that a =

y

x, b =

z

y, c =

x

z.

The Rearrangement Inequality gives

x3 + y3 + z3 ≥ x2y + y2z + z2x.


Thusa

b+

b

c+

c

a=

x3 + y3 + z3

xyz≥ x2y + y2z + z2x

xyz= a + b + c

as desired.

Third solution:Using the AM-GM Inequality, we deduce that

2a

b+

b

c≥ 3

3

√a2

bc≥ 3a.

Similarly,2b

c+

c

a≥ 3b and

2c

a+

a

b≥ 3c. Adding these three inequalities, the conclusion

is immediate.

Forth solution:

Let x = 9

√ab4

c2, y = 9

√ca4

b2, z = 9

√bc4

a2. Consequently, a = xy2, b = zx2, c = yz2,

and also xyz ≤ 1. Thus, using the Rearrangement Inequality, we find that∑ a

b=∑ x2

yz≥ xyz

∑ x2

yz=∑

x3 ≥∑

xy2 =∑

a.

15. [ Vasile Cırtoaje, Mircea Lascu ] Let a, b, c, x, y, z be positive real numberssuch that a+x ≥ b+y ≥ c+z and a+b+c = x+y+z. Prove that ay+bx ≥ ac+xz.

Solution:We have

ay+bx−ac−xz = a(y−c)+x(b−z) = a(a+b−x−z)+x(b−z) = a(a−x)+(a+x)(b−z) =

=12(a− x)2 +

12(a2 − x2) + (a + x)(b− z) =

12(a− x)2 +

12(a + x)(a− x + 2b− 2z) =

=12(a− x)2 +

12(a + x)(b− c + y − z) ≥ 0.

The equality occurs when a = x, b = z, c = y and 2x ≥ y + z.

16. [ Vasile Cırtoaje, Mircea Lascu ] Let a, b, c be positive real numbers so thatabc = 1. Prove that

1 +3

a + b + c≥ 6

ab + ac + bc.


Solution:We set x =

1a, y =

1b, z =

1c

and observe that xyz = 1. The inequality isequivalent to

1 +3

xy + yz + zx≥ 6

x + y + z.

34 Solutions

From (x + y + z)2 ≥ 3(xy + yz + zx) we get

1 +3

xy + yz + zx≥ 1 +

9(x + y + z)2

,

so it suffices to prove that

1 +9

(x + y + z)2≥ 6

x + y + z.

The last inequality is equivalent to(

1− 3x + y + z

)2

≥ 0 and this ends the proof.


a3

b2+

b3

c2+

c3

a2≥ a2

b+

b2

c+

c2

a.

JBMO 2002 Shortlist

First solution:We have

a3

b2≥ a2

b+ a− b ⇔ a3 + b3 ≥ ab(a + b) ⇔ (a− b)2(a + b) ≥ 0,

which is clearly true. Writing the analogous inequalities and adding them up gives

a3

b2+

b3

c2+

c3

a2≥ a2

b+ a− b +

b2

c+ b− c +

c2

a+ c− a =

a2

b+

b2

c+

c2

a.

Second solution:By the Cauchy-Schwarz Inequality we have

(a + b + c)(

a3

b2+

b3

c2+

c3

a2

)≥(

a2

b+

b2

c+

c2

a

)2

,

so we only have to prove that

a2

b+

b2

c+

c2

a≥ a + b + c.

But this follows immediately from the Cauchy-Schwarz Inequality.

18. Prove that if n > 3 and x1, x2, . . . , xn > 0 have product 1, then1

1 + x1 + x1x2+

11 + x2 + x2x3

+ · · ·+ 11 + xn + xnx1

> 1.

Russia, 2004

Solution:We use a similar form of the classical substitution x1 =

a2

a1, x2 =

a3

a2, . . . , xn =

a1

an.

In this case the inequality becomesa1

a1 + a2 + a3+

a2

a2 + a3 + a4+ · · ·+ an

an + a1 + a2> 1


and it is clear, because n > 3 and ai + ai+1 + ai+2 < a1 + a2 + · · ·+ an for all i.

19. [ Marian Tetiva ] Let x, y, z be positive real numbers satisfying the condition

x2 + y2 + z2 + 2xyz = 1.

Prove thata) xyz ≤ 1

8;

b) x + y + z ≤ 32;

c) xy + xz + yz ≤ 34≤ x2 + y2 + z2;

d) xy + xz + yz ≤ 12

+ 2xyz.

Solution :a) This is very simple. From the AM-GM Inequality, we have

1 = x2 + y2 + z2 + 2xyz ≥ 4 4√

2x3y3z3 ⇒ x3y3z3 ≤ 12 · 44

⇒ xyz ≤ 18.

b) Clearly, we must have x, y, z ∈ (0, 1). If we put s = x+y+z, we get immediatelyfrom the given relation

s2 − 2s + 1 = 2 (1− x) (1− y) (1− z) .

Then, again by the AM-GM Inequality (1 − x, 1 − y, 1 − z being positive), weobtain

s2 − 2s + 1 ≤ 2(

1− x + 1− y + 1− z

3

)3

= 2(

3− s

3

)3

.

After some easy calculations this yields

2s3 + 9s2 − 27 ≤ 0 ⇔ (2s− 3) (s + 3)2 ≤ 0

and the conclusion is plain.c) These inequalities are simple consequences of a) and b):

xy + xz + yz ≤ (x + y + z)2

3≤ 1

3· 94

=34

and

x2 + y2 + z2 = 1− 2xyz ≥ 1− 2 · 18

=34.

d) This is more delicate; we first notice that there are always two of the three

numbers, both greater (or both less) than12. Because of symmetry, we may assume

that x, y ≤ 12

, or x, y ≥ 12

and then

(2x− 1) (2y − 1) ≥ 0 ⇔ x + y − 2xy ≤ 12.

36 Solutions

On the other hand,

1 = x2 + y2 + z2 + 2xyz ≥ 2xy + z2 + 2xyz ⇒

⇒ 2xy (1 + z) ≤ 1− z2 ⇒ 2xy ≤ 1− z.

Now, we only have to multiply side by side the inequalities from above

x + y − 2xy ≤ 12

and

z ≤ 1− 2xy

to get the desired result:

xz + yz − 2xyz ≤ 12− xy ⇔ xy + xz + yz ≤ 1

2+ 2xyz.

It is not important in the proof, but also notice that

x + y − 2xy = xy

(1x

+1y− 2)

> 0,

because1x

and1y

are both greater than 1.

Remark.1) One can obtain some other inequalities, using

z + 2xy ≤ 1

and the two likes

y + 2xz ≤ 1, x + 2yz ≤ 1.

For example, multiplying these inequalities by z, y, x respectively and adding the newinequalities, we get

x2 + y2 + z2 + 6xyz ≤ x + y + z,

or

1 + 4xyz ≤ x + y + z.

2) If ABC is any triangle, the numbers

x = sinA

2, y = sin

B

2, z = sin

C

2satisfy the condition of this problem; conversely, if x, y, z > 0 verify

x2 + y2 + z2 + 2xyz = 1

then there is a triangle ABC so that

x = sinA

2, y = sin

B

2, z = sin

C

2.

According to this, new proofs can be given for such inequalities.


20. [ Marius Olteanu ] Let x1, x2, x3, x4, x5 ∈ R so that x1 +x2 +x3 +x4 +x5 = 0.Prove that

| cos x1|+ | cos x2|+ | cos x3|+ | cos x4|+ | cos x5| ≥ 1.

Gazeta Matematica

Solution:It is immediate to prove that

| sin(x + y)| ≤ min{| cos x|+ | cos y|, | sinx|+ | sin y|}

and| cos(x + y)| ≤ min{| sinx|+ | cos y|, | sin y|+ | cos x|}.

Thus, we infer that

1 = | cos

(5∑

i=1

xi

)| ≤ | cos x1|+| sin

(5∑

i=2

xi

)| ≤ | cos x1|+| cos x2|+| cos(x3+x4+x5)|.

But in the same manner we can prove that

| cos(x3 + x4 + x5)| ≤ | cos x3|+ | cos x4|+ | cos x5|


21. [ Florina Carlan, Marian Tetiva ] Prove that if x, y, z > 0 satisfy the condition

x + y + z = xyz

then xy + xz + yz ≥ 3 +√

x2 + 1 +√

y2 + 1 +√

z2 + 1.

First solution:We have

xyz = x + y + z ≥ 2√

xy + z ⇒ z (√

xy)2 − 2√

xy − z ≥ 0.

Because the positive root of the trinomial zt2 − 2t− z is

1 +√

1 + z2

z,

we get from here

√xy ≥ 1 +

√1 + z2

z⇔ z

√xy ≥ 1 +

√1 + z2.

Of course, we have two other similar inequalities. Then,

xy + xz + yz ≥ x√

yz + y√

xz + z√

xy ≥

≥ 3 +√

x2 + 1 +√

y2 + 1 +√

z2 + 1,

and we have both a proof of the given inequality, and a little improvement of it.

38 Solutions

Second solution:Another improvement is as follows. Start from

1x2

+1y2

+1z2≥ 1

xy+

1xz

+1yz

= 1 ⇒ x2y2 + x2z2 + y2z2 ≥ x2y2z2,

which is equivalent to

(xy + xz + yz)2 ≥ 2xyz (x + y + z) + x2y2z2 = 3 (x + y + z)2 .

Further on,

(xy + xz + yz − 3)2 = (xy + xz + yz)2 − 6 (xy + xz + yz) + 9 ≥≥ 3 (x + y + z)2 − 6 (xy + xz + yz) + 9 = 3

(x2 + y2 + z2

)+ 9,

so that

xy + xz + yz ≥ 3 +√

3 (x2 + y2 + z2) + 9.

But √3 (x2 + y2 + z2) + 9 ≥

√x2 + 1 +

√y2 + 1 +

√z2 + 1

is a consequence of the Cauchy-Schwarz Inequality and we have a second improve-ment and proof for the desired inequality:

xy + xz + yz ≥ 3 +√

3 (x2 + y2 + z2) + 9 ≥

≥ 3 +√

x2 + 1 +√

y2 + 1 +√

z2 + 1.

22. [ Laurentiu Panaitopol ] Prove that

1 + x2

1 + y + z2+

1 + y2

1 + z + x2+

1 + z2

1 + x + y2≥ 2,

for any real numbers x, y, z > −1.JBMO, 2003

Solution:

Let us observe that y ≤ 1 + y2

2and 1 + y + z2 > 0, so

1 + x2

1 + y + z2≥ 1 + x2

1 + z2 +1 + y2

2

and the similar relations. Setting a = 1 + x2, b = 1 + y2, c = 1 + z2, it is sufficient toprove that

a

2c + b+

b

2a + c+

c

2b + a≥ 1 (1)


for any a, b, c > 0. Let A = 2c + b, B = 2a + c, C = 2b + a. Then a =C + 4B − 2A

9,

b =A + 4C − 2B

9, c =

B + 4A− 2C

9and the inequality (1) is rewritten as

C

A+

A

B+

B

C+ 4

(B

A+

C

B+

A

C

)≥ 15.

Because A,B,C > 0, we have from the AM-GM Inequality that

C

A+

A

B+

B

C≥ 3 3

√A

B· B

C· C

A= 3

andB

A+

C

B+

A

C≥ 3 and the conclusion follows.

An alternative solution for (1) is by using the Cauchy-Schwarz Inequality:

a

2c + b+

b

2a + c+

c

2b + a=

a2

2ac + ab+

b2

2ab + cb+

c2

2bc + ac≥ (a + b + c)2

3(ab + bc + ca)≥ 1.

23. Let a, b, c > 0 with a + b + c = 1. Show that

a2 + b

b + c+

b2 + c

c + a+

c2 + a

a + b≥ 2.

Solution:Using the Cauchy-Schwarz Inequality, we find that

∑ a2 + b

b + c≥

(∑a2 + 1

)2

∑a2(b + c) +

∑a2 +

∑ab

.

And so it is enough to prove that(∑a2 + 1

)2

∑a2(b + c) +

∑a2 +

∑ab≥ 2 ⇔ 1 +

(∑a2)2

≥ 2∑

a2(b + c) + 2∑

ab.

The last inequality can be transformed as follows

1 +(∑

a2)2

≥ 2∑

a2(b + c) + 2∑

ab ⇔ 1 +(∑

a2)2

≥

≥ 2∑

a2 − 2∑

a3 + 2∑

ab ⇔(∑

a2)2

+ 2∑

a3 ≥∑

a2,

and it is true, because

∑a3 ≥

∑a2

3(Chebyshev’s Inequality)

and (∑a2)2

≥

∑a2

3.

40 Solutions

24. Let a, b, c ≥ 0 such that a4 + b4 + c4 ≤ 2(a2b2 + b2c2 + c2a2). Prove that

a2 + b2 + c2 ≤ 2(ab + bc + ca).

Kvant, 1988

Solution:The condition ∑

a4 ≤ 2∑

a2b2

is equivalent to

(a + b + c)(a + b− c)(b + c− a)(c + a− b) ≥ 0.

In any of the cases a = b + c, b = c + a, c = a + b, the inequality∑a2 ≤ 2

∑ab

is clear. So, suppose a + b 6= c, b + c 6= a, c + a 6= b. Because at most one of thenumbers b + c− a, c + a− b, a + b− c is negative and their product is nonnegative,all of them are positive. Thus, we may assume that

a2 < ab + ac, b2 < bc + ba, c2 < ca + cb


25. Let n ≥ 2 and x1, ..., xn be positive real numbers satisfying

1x1 + 1998

+1

x2 + 1998+ ... +

1xn + 1998

=1

1998.

Prove thatn√

x1x2...xn

n− 1≥ 1998.

Vietnam, 1998

Solution:

Let1998

1998 + xi= ai. The problem reduces to proving that for any positive real

numbers a1, a2, . . . , an such that a1 + a2 + · · ·+ an = 1 we have the inequalityn∏

i=1

(1ai− 1)≥ (n− 1)n.

This inequality can be obtained by multiplying the inequalities

1ai− 1 =

a1 + · · ·+ ai−1 + ai+1 + · · ·+ an

ai≥

≥ (n− 1) n−1

√a1 . . . ai−1ai+1 . . . an

an−1i


26. [ Marian Tetiva ] Consider positive real numbers x, y, z so that

x2 + y2 + z2 = xyz.

Prove the following inequalitiesa) xyz ≥ 27;b) xy + xz + yz ≥ 27;c) x + y + z ≥ 9;d) xy + xz + yz ≥ 2 (x + y + z) + 9.

Solution :a), b), c) Using well-known inequalities, we have

xyz = x2 + y2 + z2 ≥ 3 3√

x2y2z2 ⇒ (xyz)3 ≥ 27 (xyz)2 ,

which yields xyz ≥ 27. Then

xy + xz + yz ≥ 3 3√

(xyz)2 ≥ 3 3√

272 = 27

and

x + y + z ≥ 3 3√

xyz ≥ 3 3√

27 = 9.

d) We notice that x2 < xyz ⇒ x < yz and the likes. Consequently,

xy < yz · xz ⇒ 1 < z2 ⇒ z > 1.

Hence all the three numbers must be greater than 1. Set

a = x− 1, b = y − 1, c = z − 1.

We then have a > 0, b > 0, c > 0 and

x = a + 1, y = b + 1, z = c + 1.

Replacing these in the given condition we get

(a + 1)2 + (b + 1)2 + (c + 1)2 = (a + 1) (b + 1) (c + 1)

which is the same as

a2 + b2 + c2 + a + b + c + 2 = abc + ab + ac + bc.

If we put q = ab + ac + bc, we have

q ≤ a2 + b2 + c2,√

3q ≤ a + b + c

and

abc ≤(q

3

) 32

=(3q)

32

27.

42 Solutions

Thus

q +√

3q + 2 ≤ a2 + b2 + c2 + a + b + c + 2 =

= abc + ab + ac + bc ≤(√

3q)3

27+ q

and setting x =√

3q, we have

x + 2 ≤ x3

27⇔ (x− 6) (x + 3)2 ≥ 0.

Finally, √3q = x ≥ 6 ⇒ q = ab + ac + bc ≥ 12.

Now, recall that a = x− 1, b = y − 1, c = z − 1, so we get

(x− 1) (y − 1) + (x− 1) (z − 1) + (y − 1) (z − 1) ≥ 12 ⇒

⇒ xy + xz + yz ≥ 2 (x + y + z) + 9;

and we are done.One can also prove the stronger inequality

xy + xz + yz ≥ 4 (x + y + z)− 9.

Try it!

27. Let x, y, z be positive real numbers with sum 3. Prove that√

x +√

y +√

z ≥ xy + yz + zx.

Russia, 2002

Solution:Rewrite the inequality in the form

x2 + 2√

x + y2 + 2√

y + z2 + 2√

z ≥ x2 + y2 + z2 + 2xy + 2yz + 2zx ⇔

⇔ x2 + 2√

x + y2 + 2√

y + z2 + 2√

z ≥ 9.

Now, from the AM-GM Inequality, we have

x2 + 2√

x = x2 +√

x +√

x ≥ 3 3√

x2 · x = 3x,

y2 + 2√

y ≥ 3y, z2 + 2√

z ≥ 3z,

hencex2 + y2 + z2 + 2

(√x +

√y +

√z)≥ 3(x + y + z) ≥ 9.

28. [ D. Olteanu ] Let a, b, c be positive real numbers. Prove that

a + b

b + c· a

2a + b + c+

b + c

c + a· b

2b + c + a+

c + a

a + b· c

2c + a + b≥ 3

4.

Gazeta Matematica


Solution:We set x = a+ b, y = b+ c and z = a+ c and after a few computations we obtain

the equivalent formx

y+

y

z+

z

x+

x

x + y+

y

y + z+

z

z + x≥ 9

2.

But using the Cauchy-Schwarz Inequality,

x

y+

y

z+

z

x+

x

x + y+

y

y + z+

z

z + x≥ (x + y + z)2

xy + yz + zx+

(x + y + z)2

xy + yz + zx + x2 + y2 + z2=

2(x + y + z)4

2(xy + yz + zx)(xy + yz + zx + x2 + y2 + z2)≥ 8(x + y + z)4

(xy + yz + zx + (x + y + z)2)2


29. For any positive real numbers a, b, c show that the following inequality holdsa

b+

b

c+

c

a≥ c + a

c + b+

a + b

a + c+

b + c

b + a.

India, 2002

Solution:Let us take

a

b= x,

b

c= y,

c

a= z. Observe that

a + c

b + c=

1 + xy

1 + y= x +

1− x

1 + y.

Using similar relations, the problem reduces to proving that if xyz = 1, thenx− 1y + 1

+y − 1z + 1

+z − 1x + 1

≥ 0 ⇔

⇔ (x2 − 1)(z + 1) + (y2 − 1)(x + 1) + (z2 − 1)(y + 1) ≥ 0 ⇔⇔

∑x2z +

∑x2 ≥

∑x + 3.

But this inequality is very easy. Indeed, using the AM-GM Inequality we have∑x2z ≥ 3 and so it remains to prove that

∑x2 ≥

∑x, which follows from the

inequalities ∑x2 ≥

(∑x)2

3≥∑

x.


a3

b2 − bc + c2+

b3

c2 − ac + a2+

c3

a2 − ab + b2≥ 3(ab + bc + ca)

a + b + c.

Proposed for the Balkan Mathematical Olympiad

44 Solutions

First solution:

Since a + b + c ≥ 3(ab + bc + ca)a + b + c

, it suffices to prove that:

a3

b2 − bc + c2+

b3

c2 − ca + a2+

c3

a2 − ab + b2≥ a + b + c.

From the Cauchy-Schwartz Inequality, we get∑ a3

b2 − bc + c2≥

(∑a2)2∑

a(b2 − bc + c2).

Thus we have to show that

(a2 + b2 + c2)2 ≥ (a + b + c) ·∑

a(b2 − bc + c2).

This inequality is equivalent to

a4 + b4 + c4 + abc(a + b + c) ≥ ab(a2 + b2) + bc(b2 + c2) + ca(c2 + a2),

which is just Schur’s Inequality.Remark.The inequality

a3

b2 − bc + c2+

b3

c2 − ca + a2+

c3

a2 − ab + b2≥ a + b + c

was proposed by Vasile Cartoaje in Gazeta Matematica as a special case (n = 3)of the more general inequality

2an − bn − cn

b2 − bc + c2+

2bn − cn − an

c2 − ca + a2+

2cn − an − bn

a2 − ab + b2≥ 0.

Second solution:Rewrite our inequality as∑ (b + c)a3

b3 + c3≥ 3(ab + bc + ca)

a + b + c.

But this follows from a more general result:If a, b, c, x, y, z > 0 then

∑ a(y + z)b + c

≥ 3

∑xy∑x

.

But this inequality is an immediate (and weaker) consequence of the result fromproblem 101.

Third solution:Let

A =∑ a3

b2 − bc + c2


and

B =∑ b3 + c3

b2 − bc + c2=∑ (b + c)(b2 − bc + c2)

b2 − bc + c2= 2

∑a.

So we get

A + B =(∑

a3)(∑ 1

b2 − bc + c2

)=

=12

(∑(b + c)(b2 − bc + c2)

)(∑ 1b2 − bc + c2

)≥ 1

2

(∑√b + c

)2

,

from the Cauchy-Schwarz Inequality. Hence

A ≥ 12

(∑√b + c

)2

− 2∑

a =∑√

a + b ·√

b + c−∑

a.

We denote

Aa =√

c + a ·√

b + a− a =

∑ab√∑

ab + a2 + a

and the similar relations with Ab and Ac. So A ≥ Aa+Ab+Ac. But because(∑

a)2

≥

3(∑

ab)

we get

Aa ≥

∑ab√(∑

a)2

3 + a2 + a

=3∑

ab(∑a)2

√√√√(∑ a

)2

3+ a2 − a

and also the similar inequalities are true. So we only need to prove that

∑√√√√(∑ a

)2

3+ a2 − a

≥ a + b + c ⇔∑√

13

+(

a

a + b + c

)2

≥ 2.

We consider the convex function f(t) =

√13

+ t2. Using Jensen’s Inequality

we finally deduce that ∑√13

+(

a

a + b + c

)2

≥ 2.

We have equality if and only if a = b = c.

31. [ Adrian Zahariuc ] Consider the pairwise distinct integers x1, x2, . . . , xn,n ≥ 0. Prove that

x21 + x2

2 + · · ·+ x2n ≥ x1x2 + x2x3 + · · ·+ xnx1 + 2n− 3.

46 Solutions

Solution:The inequality can be rewritten as

n∑i=1

(xi − xi+1)2 ≥ 2(2n− 3).

Let xm = min{x1, x2, . . . , xn} and xM = max{x1, x2, . . . , xn}. Suppose without lossof generality that m < M . Let

S1 = (xm − xm+1)2 + · · ·+ (xM−1 − xM )2

and

S2 = (xM − xM+1)2 + · · ·+ (xn − x1)2 + (x1 − x2)2 + · · ·+ (xm−1 − xm)2.

The inequalityk∑

i=1

a2i ≥

1k

(k∑

i=1

ai

)2

(which follows from the Cauchy-Schwarz In-

equality) implies that

S1 ≥(xM − xm)2

M −mand

S2 ≥(xM − xm)2

n− (M −m).

Son∑

i=1

(xi − xi+1)2 = S1 + S2 ≥ (xM − xm)2(

1M −m

+1

n− (M −m)

)≥

≥ (n− 1)24n

= 4n− 8 +4n

> 4n− 8.

Butn∑

i=1

(xi − xi+1)2 ≡n∑

i=1

xi − xi+1 = 0 mod 2

son∑

i=1

(xi − xi+1)2 ≥ 4n− 6

and the problem is solved.

32. [ Murray Klamkin ] Find the maximum value of the expression x21x2 +x2

2x3 +· · ·+ x2

n−1xn + x2nx1 when x1, x2, . . . , xn−1, xn ≥ 0 add up to 1 and n > 2.

Crux Mathematicorum

Solution:First of all, it is clear that the maximum is at least

427

, because this value is

attained for x1 =13, x2 =

23, x3 = · · · = xn = 0. Now, we will prove by induction that

x21x2 + x2

2x3 + · · ·+ x2n−1xn + x2

nx1 ≤427


for all x1, x2, . . . , xn−1, xn ≥ 0 which add up to 1. Let us prove first the inductivestep. Suppose the inequality is true for n and we will prove it for n + 1. We can ofcourse assume that x2 = min{x1, x2, . . . , xn+1}. But this implies that

x21x2 + x2

2x3 + · · ·+ x2nx1 ≤ (x1 + x2)2x3 + x2

3x4 + · · ·+ x2n−1xn + x2

n(x1 + x2).

But from the inductive hypothesis we have

(x1 + x2)2x3 + x23x4 + · · ·+ x2

n−1xn + x2n(x1 + x2) ≤

427

and the inductive step is proved. Thus, it remains to prove that a2b + b2c + c2a ≤ 427

if a + b + c = 1. We may of course assume that a is the greatest among a, b, c. In this

case the inequality a2b + b2c + c2a ≤(a +

c

2

)2

·(b +

c

2

)follows immediately from

abc ≥ b2c,a2c

2≥ ac2

2. Because

1 =a +

c

22

+a +

c

22

+ b +c

2≥ 3

3

√√√√(b +c

2

)(a +

c

2

)2

4,

we have proved that a2b+ b2c+ c2a ≤ 427

and this shows that the maximum is indeed427

.

33. Find the maximum value of the constant c such that for anyx1, x2, . . . , xn, · · · > 0 for which xk+1 ≥ x1 + x2 + · · ·+ xk for any k, the inequality

√x1 +

√x2 + · · ·+

√xn ≤ c

√x1 + x2 + · · ·+ xn

also holds for any n.IMO Shortlist, 1986

Solution:First, let us see what happens if xk+1 and x1+x2+· · ·+xk are close for any k. For

example, we can take xk = 2k, because in this case we have x1+x2+· · ·+xk = xk+1−2.Thus, we find that

c ≥

n∑k=1

√2

k

√√√√ n∑k=1

2k

for any n. Taking the limit, we find that c ≥ 1 +√

2. Now, let us prove that 1 +√

2works. We will prove the inequality

√x1 +

√x2 + · · ·+

√xn ≤ (1 +

√2)√

x1 + x2 + · · ·+ xn

48 Solutions

by induction. For n = 1 or n = 2 it is clear. Suppose it is true for n and we willprove that

√x1 +

√x2 + · · ·+√xn +√xn+1 ≤ (1 +

√2)√

x1 + x2 + · · ·+ xn + xn+1.Of course, it is enough to prove that

√xn+1 ≤ (1 +

√2)(√

x1 + x2 + · · ·+ xn+1 −√

x1 + x2 + · · ·+ xn

)which is equivalent to

√x1 + x2 + · · ·+ xn +

√x1 + x2 + · · ·+ xn+1 ≤ (1 +

√2)√

xn+1.

But this one follows because

x1 + x2 + · · ·+ xn ≤ xn+1.

34. Given are positive real numbers a, b, c and x, y, z, for which a + x = b + y =c + z = 1. Prove that

(abc + xyz)( 1

ay+

1bz

+1cx

)≥ 3.

Russia, 2002

Solution:Let us observe that abc + xyz = (1− b)(1− c) + ac + ab− a. Thus,

1− c

a+

c

1− b− 1 =

abc + xyz

a(1− b).

Using these identities we deduce immediately that

3 + (xyz + abc)(

1ay

+1bz

+1cx

)=

a

1− c+

b

1− a+

c

1− b+

1− c

a+

1− a

b+

1− b

c.

Now, all we have to do is apply the AM-GM Inequalitya

1− c+

b

1− a+

c

1− b+

1− c

a+

1− a

b+

1− b

c≥ 6.

35. [ Viorel Vajaitu, Alexandru Zaharescu ] Let a, b, c be positive real numbers.Prove that

ab

a + b + 2c+

bc

b + c + 2a+

ca

c + a + 2b≤ 1

4(a + b + c).

Gazeta Matematica

First solution:We have the following chain of inequalities∑ ab

a + b + 2c=∑ ab

a + c + b + c≤∑ ab

4

(1

a + c+

1b + c

)=

a + b + c

4.


Second solution:Because the inequality is homogeneous we can consider without loss of generality

that a + b + c = 1 and so the inequality is equivalent to∑ 1a(a + 1)

≤ 14abc

.

We have1

t(t + 1)=

1t− 1

t + 1, so the inequality is equivalent to

∑ 1a≤∑ 1

a + 1+

14abc

.

We will prove now the following intercalation:∑ 1a≤ 9

4+

14abc

≤∑ 1

a + 1+

14abc

.

The inequality in the right follows from the Cauchy-Schwarz Inequality:(∑ 1a + 1

)(∑(a + 1)

)≥ 9

and the identity∑

(a + 1) = 4. The inequality in the left can be written as∑

ab ≤1 + 9abc

4, which is exactly Schur’s Inequality.

36. Find the maximum value of the expression

a3(b + c + d) + b3(c + d + a) + c3(d + a + b) + d3(a + b + c)

where a, b, c, d are real numbers whose sum of squares is 1.

Solution:The idea is to observe that a3(b+c+d)+b3(c+d+a)+c3(d+a+b)+d3(a+b+c) is

equal to∑

ab(a2+b2). Now, because the expression ab(a2+b2) appears when writing(a− b)4, let us see how the initial expression can be written:∑

ab(a2 + b2) =∑ a4 + b4 + 6a2b2 − (a− b)4

4=

=3∑

a4 + 6∑

a2b2 −∑

(a− b)4

4=

3−∑

(a− b)4

4≤ 3

4.

The maximum is attained for a = b = c = d =12.

37. [ Walther Janous ] Let x, y, z be positive real numbers. Prove thatx

x +√

(x + y)(x + z)+

y

y +√

(y + z)(y + x)+

z

z +√

(z + x)(z + y)≤ 1.

Crux Mathematicorum

50 Solutions

First solution:We have (x+y)(x+z) = xy+(x2 +yz)+xz ≥ xy+2x

√yz+xz = (

√xy+

√xz)2.

Hence ∑ x

x +√

(x + y)(x + z)≤∑ x

x +√

xy +√

xz.

But ∑ x

x +√

xy +√

xz=∑ √

x√x +

√y +

√z

= 1

and this solves the problem.

Second solution:From Huygens Inequality we have

√(x + y)(x + z) ≥ x +

√yz and using this

inequality for the similar ones we getx

x +√

(x + y)(x + z)+

y

y +√

(y + z)(y + x)+

z

z +√

(z + x)(z + y)≤

≤ x

2x +√

yz+

y

2y +√

zx+

z

2z +√

xy.

Now, we denote with a =√

yz

x, b =

√zx

y, c =

√xy

zand the inequality becomes

12 + a

+1

2 + b+

12 + c

≤ 1.

From the above notations we can see that abc = 1, so the last inequality becomesafter clearing the denominators ab + bc + ca ≥ 3, which follows from the AM-GMInequality.

38. Suppose that a1 < a2 < ... < an are real numbers for some integer n ≥ 2.Prove that

a1a42 + a2a

43 + ... + ana4

1 ≥ a2a41 + a3a

42 + ... + a1a

4n.

Iran, 1999

Solution:A quick look shows that as soon as we prove the inequality for n = 3, it will be

proved by induction for larger n. Thus, we must prove that for any a < b < c we haveab(b3 − a3) + bc(c3 − b3) ≥ ca(c3 − a3) ⇔ (c3 − b3)(ac − bc) ≤ (b3 − a3)(ab − ac).Because a < b < c, the last inequality reduces to a(b2 + ab + a2) ≤ c(c2 + bc + b2).And this last inequality is equivalent to (c− a)(a2 + b2 + c2 + ab+ bc+ ca) ≥ 0, whichis clear.

39. [ Mircea Lascu ] Let a, b, c be positive real numbers. Prove that

b + c

a+

c + a

b+

a + b

c≥ 4

(a

b + c+

b

c + a+

c

a + b

).


Solution:Using the inequality

1x + y

≤ 14x

+14y

we infer that

4a

b + c≤ a

b+

a

c,

4b

a + c≤ b

a+

b

cand

4c

a + b≤ c

a+

c

b.

Adding up these three inequalities, the conclusion follows.

40. Let a1, a2, . . . , an > 1 be positive integers. Prove that at least one of thenumbers a1

√a2, a2

√a3, . . . , an−1

√an, an

√a1 is less than or equal to 3

√3.

Adapted after a well-known problem

Solution:Suppose we have a

1a1i+1 > 3

13 for all i. First, we will prove that n

1n ≤ 3

13 for all

natural number n. For n = 1, 2, 3, 4 it is clear. Suppose the inequality is true for n > 3and let us prove it for n + 1. This follows from the fact that

1 +1n≤ 1 +

14

<3√

3 ⇒ 3n+1

3 = 3√

3 · 3n3 ≥ n + 1

n· n = n + 1.

Thus, using this observation, we find that a1

aii+1 > 3

13 ≥ a

1ai+1i+1 ⇒ ai+1 > ai for all i,

which means that a1 < a2 < · · · < an−1 < an < a1, contradiction.

41. [ Mircea Lascu, Marian Tetiva ] Let x, y, z be positive real numbers whichsatisfy the condition

xy + xz + yz + 2xyz = 1.

Prove that the following inequalities hold

a) xyz ≤ 18;

b) x + y + z ≥ 32;

c)1x

+1y

+1z≥ 4 (x + y + z);

d)1x

+1y

+1z− 4 (x + y + z) ≥ (2z − 1)2

(z (2z + 1)), where z = max {x, y, z}.

Solution:a) We put t3 = xyz; according to the AM-GM Inequality we have

1 = xy + xz + yz + 2xyz ≥ 3t2 + 2t3 ⇔ (2t− 1) (t + 1)2 ≤ 0,

therefore 2t− 1 ≤ 0 ⇔ t ≤ 12, this meaning that xyz ≤ 1

8.

b) Denote also s = x + y + z; the following inequalities are well-known

(x + y + z)2 ≥ 3 (xy + xz + yz)

and(x + y + z)3 ≥ 27xyz;

52 Solutions

then we have 2s3 ≥ 54xyz = 27− 27 (xy + xz + yz) ≥ 27− 9s2, i. e.

2s3 + 9s2 − 27 ≥ 0 ⇔ (2s− 3) (s + 3)2 ≥ 0,

where from 2s− 3 ≥ 0 ⇔ s ≥ 32.

Or, because p ≤ 18, we have

s2 ≥ 3q = 3 (1− 2p) ≥ 3(

1− 28

)=

94;

if we put q = xy + xz + yz, p = xyz.Now, one can see the following is also true

q = xy + xz + yz ≥ 34.

c) The three numbers x, y, z cannot be all less than12, because, in this case we

get the contradiction

xy + xz + yz + 2xyz <34

+ 2 · 18

= 1;

because of symmetry we may assume then that z ≥ 12.

We have 1 = (2z + 1) xy + z (x + y) ≥ (2z + 1) xy + 2z√

xy, which can alsobe written in the form

((2z + 1)

√xy − 1

) (√xy + 1

)≤ 0; and this one yields the

inequality

xy ≤ 1(2z + 1)2

.

We also have 1 = (2z + 1)xy + z (x + y) ≤ (2z + 1)(x + y)2

4+ z (x + y), conse-

quently ((2z + 1) (x + y)− 2) (x + y + 2) ≥ 0, which shows us that

x + y ≥ 22z + 1

.

The inequality to be proved

1x

+1y

+1z≥ 4 (x + y + z)

can be also rearranged as

(x + y)(

1xy− 4)≥ 4z2 − 1

z=

(2z − 1) (2z + 1)z

.

From the above calculations we infer that

(x + y)(

1xy− 4)≥ 2

2z + 1

((2z + 1)2 − 4

)=

2 (2z − 1) (2z + 3)2z + 1


(the assumption z ≥ 12

allows the multiplication of the inequalities side by side), andthis means that the problem would be solved if we proved

2 (2z + 3)2z + 1

≥ 2z + 1z

⇔ 4z2 + 6z ≥ 4z2 + 4z + 1;

but this follows for z ≥ 12

and we are done.

d) Of course, if z is the greatest from the numbers x, y, z, then z ≥ 12; we saw

that

1x

+1y− 4 (x + y) = (x + y)

(1xy− 4)≥ 2

2z + 1(2z + 1)

=2 (2z − 1) (2z + 3)

2z + 1= 4z − 1

z+

(2z − 1)2

z (2z + 1),

where from we get our last inequality

1x

+1y

+1z− 4 (x + y + z) ≥ (2z − 1)2

z (2z + 1).

Of course, in the right-hand side z could be replaced by any of the three numbers

which is ≥ 12

(two such numbers could be, surely there is one).

Remark.It is easy to see that the given condition implies the existence of positive numbers

a, b, c such that x =a

b + c, y =

b

c + a, z =

c

a + b. And now a), b) and c) reduce

immediately to well-known inequalities! Try to prove using this substitution d).

42. [ Manlio Marangelli ] Prove that for any positive real numbers x, y, z,

3(x2y + y2z + z2x)(xy2 + yz2 + zx2) ≥ xyz(x + y + z)3.

Solution:Using the AM-GM Inequality, we find that

13+

y2z

y2z + z2x + x2y+

xy2

yz2 + zx2 + xy2≥

3y 3√

xyz3√

3 · (y2z + z2x + x2y) · (yz2 + zx2 + xy2)

and two other similar relations

13+

z2x

y2z + z2x + x2y+

yz2

yz2 + zx2 + xy2≥

3z 3√

xyz3√

3 · (y2z + z2x + x2y) · (yz2 + zx2 + xy2)

13+

x2y

y2z + z2x + x2y+

zx2

yz2 + zx2 + xy2≥

3x 3√

xyz3√

3 · (y2z + z2x + x2y) · (yz2 + zx2 + xy2)Then, adding up the three relations, we find exactly the desired inequality.

54 Solutions

43. [ Gabriel Dospinescu ] Prove that if a, b, c are real numbers such thatmax{a, b, c} −min{a, b, c} ≤ 1, then

1 + a3 + b3 + c3 + 6abc ≥ 3a2b + 3b2c + 3c2a

Solution:

Clearly, we may assume that a = min{a, b, c} and let us write b = a+x, c = a+y,where x, y ∈ [0, 1]. It is easy to see that a3 +b3 +c3−3abc = 3a(x2−xy+y2)+x3 +y3

and a2b + b2c + c2a − 3abc = a(x2 − xy + y2) + x2y. So, the inequality becomes1 + x3 + y3 ≥ 3x2y. But this follows from the fact that 1 + x3 + y3 ≥ 3xy ≥ 3x2y,because 0 ≤ x, y ≤ 1.

44. [ Gabriel Dospinescu ] Prove that for any positive real numbers a, b, c we have

27 +(

2 +a2

bc

)(2 +

b2

ca

)(2 +

c2

ab

)≥ 6(a + b + c)

(1a

+1b

+1c

).

Solution:By expanding the two sides, the inequality is equivalent to 2abc(a3 + b3 + c3 +

3abc − a2b − a2c − b2a − b2c − c2a − c2b) + (a3b3 + b3c3 + c3a3 + 3a2b2c2 − a3b2c −a3bc2− ab3c2− ab2c3− a2b3c− a2bc3) ≥ 0. But this is true from Schur’s Inequalityapplied for a, b, c and ab, bc, ca.

45. Let a0 =12

and ak+1 = ak +a2

k

n. Prove that 1− 1

n< an < 1.

TST Singapore

Solution:We have

1ak+1

=1ak− 1

ak + nand so

n−1∑k=0

(1ak− 1

ak+1

)=

n−1∑k=0

1n + ak

< 1 ⇒ 2− 1an

< 1 ⇒ an < 1.

Now, because the sequence is increasing we also have 2− 1an

=n−1∑k=0

1n + ak

>n

n + 1and from this inequality and the previous one we conclude immediately that

1− 1n

< an < 1.

46. [ Calin Popa ] Let a, b, c be positive real numbers, with a, b, c ∈ (0, 1) suchthat ab + bc + ca = 1. Prove that

a

1− a2+

b

1− b2+

c

1− c2≥ 3

4

(1− a2

a+

1− b2

b+

1− c2

c

).


Solution:It is known that in every triangle ABC the following identity holds∑tan

A

2tan

B

2= 1 and because tan is bijective on

[0,

π

2

)we can set a = tan

A

2, b =

tanB

2, c = tan

C

2. The condition a, b, c ∈ (0, 1) tells us that the triangle ABC is

acute-angled. With this substitution, the inequality becomes

∑ 2 tanA

2

1− tan2 A

2

≥ 3∑ 1− tan2 A

2

2 tanA

2

⇔∑

tanA ≥ 3∑ 1

tanA⇔

⇔ tanA tanB tanC(tanA+tanB+tanC) ≥ 3(tanA tanB+tanB tanC+tanC tanA) ⇔

⇔ (tanA + tanB + tanC)2 ≥ 3(tanA tanB + tanB tanC + tanC tanA),

clearly true because tanA, tanB, tanC > 0.

47. [ Titu Andreescu, Gabriel Dospinescu ] Let x, y, z ≤ 1 and x + y + z = 1.Prove that

11 + x2

+1

1 + y2+

11 + z2

≤ 2710

.

Solution:Using the fact that (4− 3t)(1− 3t)2 ≥ 0 for any t ≤ 1, we find that

11 + x2

≤ 2750

(2− x).

Writing two similar expressions for y and z and adding them up, we find the desiredinequality.

Remark.Tough it may seem too easy, this problem helps us to prove the following difficult

inequality ∑ (b + c− a)2

a2 + (b + c)2≥ 3

5.

In fact, this problem is equivalent to that difficult one. Try to prove this!

48. [ Gabriel Dospinescu ] Prove that if√

x +√

y +√

z = 1, then

(1− x)2(1− y)2(1− z)2 ≥ 215xyz(x + y)(y + z)(z + x)

.

Solution:We put a =

√x, b =

√y and c =

√z. Then 1− x = 1− a2 = (a + b + c)2 − a2 =

(b + c)(2a + b + c). Now we have to prove that ((a + b)(b + c)(c + a)(2a + b + c)(a +2b + c)(a + b + 2c))2 ≥ 215a2b2c2(a2 + b2)(b2 + c2)(c2 + a2). But this inequality is true

56 Solutions

as it follows from the following true inequalities ab(a2 + b2) ≤ (a + b)4

8(this being

equivalent to (a−b)4 ≥ 0) and (2a+b+c)(a+2b+c)(a+b+2c) ≥ 8(b+c)(c+a)(a+b).

49. Let x, y, z be positive real numbers such that xyz = x+ y + z +2. Prove that

(1) xy + yz + zx ≥ 2(x + y + z);

(2)√

x +√

y +√

z ≤ 32√

xyz.

Solution:The initial condition xyz = x + y + z + 2 can be rewritten as follows

11 + x

+1

1 + y+

11 + z

= 1.

Now, let1

1 + x= a,

11 + y

= b,1

1 + z= c.

Thenx =

1− a

a=

b + c

a, y =

c + a

b, z =

a + b

c.

(1) We have

xy + yz + zx ≥ 2(x + y + z) ⇔ b + c

a· c + a

b+

c + a

b· a + b

c+

a + b

c· b + c

a≥

≥ 2(

b + c

a+ ·c + a

b+

a + b

c

)⇔ a3 + b3 + c3 + 3abc ≥

≥ ab(a + b) + bc(b + c) + ca(c + a) ⇔∑

a(a− b)(a− c) ≥ 0,

which is exactly Schur’s Inequality.(2) Here we have

√x +

√y +

√z ≤ 3

2√

xyz ⇔

⇔√

a

b + c· b

c + a+

√b

c + a· c

a + b+√

c

a + b· a

b + c≤ 3

2.

This can be proved by adding the inequality√a

b + c· b

c + a≤ 1

2

(a

a + c+

b

b + c

),

with the analogous ones.

50. Prove that if x, y, z are real numbers such that x2 + y2 + z2 = 2, then

x + y + z ≤ xyz + 2.

IMO Shortlist, 1987


First solution:If one of x, y, z is negative, let us say x then

2 + xyz − x− y − z = (2− y − z)− x(1− yz) ≥ 0,

because y + z ≤√

2(y2 + z2) ≤ 2 and zy ≤ z2 + y2

2≤ 1. So we may assume that

0 < x ≤ y ≤ z. If z ≤ 1 then

2 + xyz − x− y − z = (1− z)(1− xy) + (1− x)(1− y) ≥ 0.

Now, if z > 1 we have

z + (x + y) ≤√

2(z2 + (x + y)2) = 2√

1 + xy ≤ 2 + xy ≤ 2 + xyz.

This ends the proof.

Second solution:Using the Cauchy-Schwarz Inequality, we find that

x + y + z − xyz = x(1− yz) + y + z ≤√

(x2 + (y + z)2) · (1 + (1− yz)2).

So, it is enough to prove that this last quantity is at most 2, which is equivalent tothe inequality (2 + 2yz)(2 − 2yz + (yz)2) ≤ 4 ⇔ 2(yz)3 ≤ 2(yz)2, which is clearlytrue, because 2 ≥ y2 + z2 ≥ 2yz.

51. [ Titu Andreescu, Gabriel Dospinescu ] Prove that for any x1, x2, . . . , xn ∈(0, 1) and for any permutation σ of the set {1, 2, . . . , n}, we have the inequality

n∑i=1

11− xi

≥

1 +

n∑i=1

xi

n

·(

n∑i=1

11− xi · xσ(i)

).

Solution:Using the AM-GM Inequality and the fact that

1x + y

≤ 14x

+14y

, we can

write the following chain of inequalities

n∑i=1

11− xi · yi

≤n∑

i=1

1

1− x2i + y2

i

2

= 2n∑

i=1

11− x2

i + 1− y2i

≤

≤n∑

i=1

(1

2(1− x2i )

+1

2(1− y2i )

)=

n∑i=1

11− x2

i

.

58 Solutions

So, it is enough to prove the inequality

n∑i=1

11− xi

≥

1 +

n∑i=1

xi

n

·(

n∑i=1

11− x2

i

)⇔

⇔n∑

i=1

xi

1− x2i

≥ 1n·

(n∑

i=1

xi

)·

(n∑

i=1

11− x2

i

),

which is Chebyshev Inequality for the systems (x1, x2, . . . , xn) and(1

1− x21

,1

1− x22

, . . . ,1

1− x2n

).

52. Let x1, x2, . . . , xn be positive real numbers such thatn∑

i=1

11 + xi

= 1. Prove

thatn∑

i=1

√xi ≥ (n− 1)

n∑i=1

1√

xi.

Vojtech Jarnik

First solution:

Let1

1 + xi= ai. The inequality becomes

n∑i=1

√1− ai

ai≥ (n− 1)

n∑i=1

√ai

1− ai⇔

n∑i=1

1√ai(1− ai)

≥

≥ nn∑

i=1

√ai

1− ai⇔ n

n∑i=1

√ai

1− ai≤

(n∑

i=1

ai

)(n∑

i=1

1√ai(1− ai)

).

But the last inequality is a consequence of Chebyshev’s Inequality for the n-tuples(a1, a2, . . . , an) and(

1√a1(1− a1)

,1√

a2(1− a2), . . . ,

1√an(1− an)

).

Solution 2 :With the same notations, we have to prove that

(n− 1)n∑

i=1

√ai

a1 + a2 + . . . + ai−1 + ai+1 + . . . + an≤

≤n∑

i=1

√a1 + a2 + . . . + ai−1 + ai+1 + . . . + an

ai.


But using the Cauchy-Schwarz Inequality and the AM-GM Inequality, wededuce that

n∑i=1

√a1 + a2 + . . . + ai−1 + ai+1 + . . . + an

ai≥

≥n∑

i=1

√a1 +

√a2 + . . . +√

ai−1 +√ai+1 + . . . +

√an√

n− 1 · √ai

=

=n∑

i=1

√ai√

n− 1

(1√

a1+

1√

a2+ · · ·+ 1

√ai−1

+1

√ai+1

+ · · ·+ 1√

an

)≥

≥n∑

i=1

(n− 1)√

n− 1 · √ai√a1 +

√a2 + · · ·+√

ai−1 +√ai+1 + · · ·+√

an≥

≥n∑

i=1

(n− 1)√

ai

a1 + a2 + · · ·+ ai−1 + ai+1 + · · ·+ an

and we are done.

53. [ Titu Andreescu ] Let n > 3 and a1, a2, . . . , an be real numbers such thata1 +a2 + . . .+an ≥ n and a2

1 +a22 + . . .+a2

n ≥ n2. Prove that max{a1, a2, . . . , an} ≥ 2.

USAMO, 1999

Solution:The most natural idea is to suppose that ai < 2 for all i and to substitute

xi = 2− ai > 0. Then we haven∑

i=1

(2− xi) ≥ n ⇒n∑

i=1

xi ≤ n and also

n2 ≤n∑

i=1

a2i =

n∑i=1

(2− xi)2 = 4n− 4n∑

i=1

xi +n∑

i=1

x2i

Now, using the fact that xi > 0, we obtainn∑

i=1

x2i <

(n∑

i=1

xi

)2

, which combined with

the above inequality yields

n2 < 4n− 4n∑

i=1

xi +

(n∑

i=1

xi

)2

< 4n + (n− 4)n∑

i=1

xi

(we have used the fact thatn∑

i=1

xi ≤ n). Thus, we have (n − 4)

(n∑

i=1

xi − n

)> 0,

which is clearly impossible since n ≥ 4 andn∑

i=1

xi ≤ n. So, our assumption was wrong

and consequently max{a1, a2, . . . , an} ≥ 2.

60 Solutions

54. [ Vasile Cırtoaje ] If a, b, c, d are positive real numbers, then

a− b

b + c+

b− c

c + d+

c− d

d + a+

d− a

a + b≥ 0.

Solution:We have

a− b

b + c+

b− c

c + d+

c− d

d + a+

d− a

a + b=

a + c

b + c+

b + d

c + d+

c + a

d + a+

d + b

a + b− 4 =

= (a + c)(

1b + c

+1

d + a

)+ (b + d)

(1

c + d+

1a + b

)− 4.

Since1

b + c+

1d + a

≥ 4(b + c) + (d + a)

,1

c + d+

1a + b

≥ 4(c + d) + (a + b)

,

we get

a− b

b + c+

b− c

c + d+

c− d

d + a+

d− a

a + b≥ 4(a + c)

(b + c) + (d + a)+

4(b + d)(c + d) + (a + b)

− 4 = 0.

Equality holds for a = c and b = d.

Conjecture (Vasile Cırtoaje)If a, b, c, d, e are positive real numbers, then

a− b

b + c+

b− c

c + d+

c− d

d + e+

d− e

e + a+

e− a

a + b≥ 0.

55. If x and y are positive real numbers, show that xy + yx > 1.France, 1996

Solution:We will prove that ab ≥ a

a + b− abfor any a, b ∈ (0, 1). Indeed, from Bernoulli

Inequality it follows that a1−b = (1 + a − 1)1−b ≤ 1 + (a − 1)(1 − b) = a + b − ab

and thus the conclusion. Now, it x or y is at least 1, we are done. Otherwise, let0 < x, y < 1. In this case we apply the above observation and find that xy + yx ≥

x

x + y − xy+

y

x + y − xy>

x

x + y+

y

x + y= 1.

56. Prove that if a, b, c > 0 have product 1, then

(a + b)(b + c)(c + a) ≥ 4(a + b + c− 1).

MOSP, 2001


First solution:Using the identity (a + b)(b + c)(c + a) = (a + b + c)(ab + bc + ca)− 1 we reduce

the problem to the following one

ab + bc + ca +3

a + b + c≥ 4.

Now, we can apply the AM-GM Inequality in the following form

ab + bc + ca +3

a + b + c≥ 4 4

√(ab + bc + ca)3

9(a + b + c).

And so it is enough to prove that

(ab + bc + ca)3 ≥ 9(a + b + c).

But this is easy, because we clearly have ab + bc + ca ≥ 3 and (ab + bc + ca)2 ≥3abc(a + b + c) = 3(a + b + c).

Second solution:We will use the fact that (a + b)(b + c)(c + a) ≥ 8

9(a + b + c)(ab + bc + ca).

So, it is enough to prove that29(ab + bc + ca) +

1a + b + c

≥ 1. Using the AM-GM

Inequality, we can write

29(ab + bc + ca) +

1a + b + c

≥ 3 3

√(ab + bc + ca)2

81(a + b + c)≥ 1,

because

(ab + bc + ca)2 ≥ 3abc(a + b + c) = 3(a + b + c).

57. Prove that for any a, b, c > 0,

(a2 + b2 + c2)(a + b− c)(b + c− a)(c + a− b) ≤ abc(ab + bc + ca).

Solution:Clearly, if one of the factors in the left-hand side is negative, we are done. So,

we may assume that a, b, c are the side lenghts of a triangle ABC. With the usualnotations in a triangle, the inequality becomes

(a2+b2+c2)· 16K2

a + b + c≤ abc(ab+bc+ca) ⇔ (a+b+c)(ab+bc+ca)R2 ≥ abc(a2+b2+c2).

But this follows from the fact that (a + b + c)(ab + bc + ca) ≥ 9abc and

0 ≤ OH2 = 9R2 − a2 − b2 − c2.

62 Solutions

58. [ D.P.Mavlo ] Let a, b, c > 0. Prove that

3 + a + b + c +1a

+1b

+1c

+a

b+

b

c+

c

a≥ 3

(a + 1)(b + 1)(c + 1)1 + abc

.

Kvant, 1988

Solution:The inequality is equivalent to the following one

∑a +

∑ 1a

+∑ a

b≥ 3

∑ab +

∑a

abc + 1or

abc∑

a +∑ 1

a+∑

a2c +∑ a

b≥ 2

(∑a +

∑ab)

.

But this follows from the inequalities

a2bc +b

c≥ 2ab, b2ca +

c

a≥ 2bc, c2ab +

a

b≥ 2ca

and

a2c +1c≥ 2a, b2a +

1a≥ 2b, c2b +

1b≥ 2c.

59. [ Gabriel Dospinescu ] Prove that for any positive real numbers x1, x2, . . . , xn

with product 1 we have the inequality

nn ·n∏

i=1

(xni + 1) ≥

(n∑

i=1

xi +n∑

i=1

1xi

)n

.

Solution:Using the AM-GM Inequality, we deduce that

xn1

1 + xn1

+xn

2

1 + xn2

+ · · ·+xn

n−1

1 + xnn−1

+1

1 + xnn

≥ n

xn · n

√√√√ n∏i=1

(1 + xni )

and1

1 + xn1

+1

1 + xn2

+ · · ·+ 11 + xn

n−1

+xn

n

1 + xnn

≥ n · xn

n

√√√√ n∏i=1

(1 + xni )

Thus, we have

n

√√√√ n∏i=1

(1 + xni ) ≥ xn +

1xn

.


Of course, this is true for any other variable, so we can add all these inequalities toobtain that

n n

√√√√ n∏i=1

(1 + xni ) ≥

n∑i=1

xi +n∑

i=1

1xi

which is the desired inequality.

60. Let a, b, c, d > 0 such that a + b + c = 1. Prove that

a3 + b3 + c3 + abcd ≥ min{

14,19

+d

27

}.

Kvant, 1993

Solution:Suppose the inequality is false. Then we have, taking into account that abc ≤ 1

27,

the inequality d

(127− abc

)> a3 + b3 + c3− 1

9. We may assume that abc <

127

. Now,

we will reach a contradiction proving that a3 + b3 + c3 + abcd ≥ 14. It is sufficient to

prove that

a3 + b3 + c3 − 19

127− abc

abc + a3 + b3 + c3 ≥ 14.

But this inequality is equivalent to 4∑

a3 + 15abc ≥ 1. We use now the identity∑a3 = 3abc+1−3

∑ab and reduce the problem to proving that

∑ab ≤ 1 + 9abc

4,

which is Schur’s Inequality.

61. Prove that for any real numbers a, b, c we have the inequality∑(1 + a2)2(1 + b2)2(a− c)2(b− c)2 ≥ (1 + a2)(1 + b2)(1 + c2)(a− b)2(b− c)2(c− a)2.

AMM

Solution:

The inequality can be also written as∑ (1 + a2)(1 + b2)

(1 + c2)(a− b)2≥ 1 (of course, we may

assume that a, b, c are distinct). Now, adding the inequalities

(1 + a2)(1 + b2)(1 + c2)(a− b)2

+(1 + b2)(1 + c2)(1 + a2)(b− c)2

≥ 21 + b2

|a− b||c− b|

(which can be found using the AM-GM Inequality) we deduce that∑ (1 + a2)(1 + b2)(1 + c2)(a− b)2

≥∑ 1 + b2

|(b− a)(b− c)|

64 Solutions

and so it is enough to prove that the last quantity is at least 1. But it follows from∑ 1 + b2

|(b− a)(b− c)|≥∣∣∣∣∑ 1 + b2

(b− a)(b− c)

∣∣∣∣ = 1

and the problem is solved.

62. [ Titu Andreescu, Mircea Lascu ] Let α, x, y, z be positive real numbers suchthat xyz = 1 and α ≥ 1. Prove that

xα

y + z+

yα

z + x+

zα

x + y≥ 3

2.

First solution:We may of course assume that x ≥ y ≥ z. Then we have

x

y + z≥ y

z + x≥ z

x + y

and xα−1 ≥ yα−1 ≥ zα−1. Using Chebyshev’s Inequality we infer that∑ xα

y + z≥ 1

3·(∑

xα−1)·(∑ x

y + z

).

Now, all we have to do is to observe that this follows from the inequalities∑

xα−1 ≥ 3

(from the AM-GM Inequality) and∑ x

y + z≥ 3

2

Second solution:According to the Cauchy-Schwarz Inequality, we have:

[(x(y+z)+y(z+x)+z(x+y)]](

xα

y + z+

yα

z + x+

zα

x + y

)≥(x

1+α2 + y

1+α2 + z

1+α2

)2

.

Thus it remains to show that(x

1+α2 + y

1+α2 + z

1+α2

)2

≥ 3(xy + yz + zx).

Since (x + y + z)2 ≥ 3(xy + yz + zx), it is enough to prove that

x1+α

2 + y1+α

2 + z1+α

2 ≥ x + y + z.

From Bernoulli’s Inequality, we get

x1+α

2 = [1 + (x− 1)]1+α

2 ≥ 1 +1 + α

2(x− 1) =

1− α

2+

1 + α

2x

and, similarly,

y1+α

2 ≥ 1− α

2+

1 + α

2y, z

1+α2 ≥ 1− α

2+

1 + α

2z.

Thus

x1+α

2 + y1+α

2 + z1+α

2 − (x + y + z) ≥ 3(1− α)2

+1 + α

2(x + y + z)− (x + y + z) =


α− 12

(x + y + z − 3) ≥ α− 12

(3 3√

xyz − 3) = 0.

Equality holds for x = y = z = 1.

Remark :Using the substitution β = α + 1 (β ≥ 2) and x =

1a, y =

1b, z =

1c

(abc = 1) theinequality becomes as follows

1aβ(b + c)

+1

bβ(c + a)+

1cγ(a + b)

≥ 32.

For β = 3, we obtain one of the problems of IMO 1995 (proposed by Russia).

63. Prove that for any real numbers x1, . . . , xn, y1, . . . , yn such that x21+· · ·+x2

n =y21 + · · ·+ y2

n = 1,

(x1y2 − x2y1)2 ≤ 2

(1−

n∑k=1

xkyk

).

Korea, 2001

Solution:We clearly have the inequality

(x1y2 − x2y1)2 ≤∑

1≤i<j≤n

(xiyj − xjyi)2 =

(n∑

i=1

x2i

)(n∑

i=1

y2i

)−

(n∑

i=1

xiyi

)2

=

=

(1−

n∑i=1

xiyi

)(1 +

n∑i=1

xiyi

).

Because we also have

∣∣∣∣∣n∑

i=1

xiyi

∣∣∣∣∣ ≤ 1, we find immediately that

(1−

n∑i=1

xiyi

)(1 +

n∑i=1

xiyi

)≤ 2

(1−

n∑i=1

xiyi

)and the problem is solved.

64. [ Laurentiu Panaitopol ] Let a1, a2, . . . , an be pairwise distinct positive inte-gers. Prove that

a21 + a2

2 + · · ·+ a2n ≥

2n + 13

(a1 + a2 + · · ·+ an).

TST Romania

66 Solutions

Solution:Without loss of generality, we may assume that a1 < a2 < · · · < an and hence

ai ≥ i for all i. Thus, we may take bi = ai − i ≥ 0 and the inequality becomesn∑

i=1

b2i + 2

n∑i=1

ibi +n(n + 1)(2n + 1)

6≥ 2n + 1

3·

n∑i=1

bi +n(n + 1)(2n + 1)

6.

Now, using the fact that ai+1 > ai we infer that b1 ≤ b2 ≤ · · · ≤ bn and fromChebyshev’s Inequality we deduce that

2n∑

i=1

ibi ≥ (n + 1)n∑

i=1

bi ≥2n + 1

3

n∑i=1

bi

and the conclusion is immediate. Also, from the above relations we can see imme-diately that we have equality if and only if a1, a2, . . . , an is a permutation of thenumbers 1, 2, . . . , n.

65. [ Calin Popa ] Let a, b, c be positive real numbers such that a + b + c = 1.Prove that

b√

c

a(√

3c +√

ab)+

c√

a

b(√

3a +√

bc)+

a√

b

c(√

3b +√

ca)≥ 3

√3

4.

Solution:Rewrite the inequality in the form

∑ √bc

a√3ca

b+ a

≥ 3√

34

.

With the substitution x =

√bc

a, y =

√ca

b, z =

√ab

cthe condition a + b + c = 1

becomes xy + yz + zx = 1 and the inequality turns into∑ x√3y + yz

≥ 3√

34

.

But, by applying the Cauchy-Schwarz Inequality we obtain

∑ x2

√3xy + xyz

≥

(∑x)2

√3 + 3xyz

≥3∑

xy

√3 +

1√3

=3√

34

,

where we used the inequalities(∑x)2

≥ 3(∑

xy)

and xyz ≤ 13√

3.


66. [ Titu Andreescu, Gabriel Dospinescu ] Let a, b, c, d be real numbers such that(1 + a2)(1 + b2)(1 + c2)(1 + d2) = 16. Prove that

−3 ≤ ab + bc + cd + da + ac + bd− abcd ≤ 5.

Solution:Let us write the condition in the form 16 =

∏(i+a) ·

∏(a− i). Using symmetric

sums, we can write this as follows

16 =(1− i

∑a−

∑ab + i

∑abc + abcd

)(1 + i

∑a−

∑ab− i

∑abc + abcd

).

So, we have the identity 16 = (1−∑

ab + abcd)2 + (∑

a−∑

abc)2. This means that|1−

∑ab + abcd| ≤ 4 and from here the conclusion follows.

67. Prove that

(a2 + 2)(b2 + 2)(c2 + 2) ≥ 9(ab + bc + ca)

for any positive real numbers a, b, c.APMO, 2004

First solution:We will prove even more: (a2 + 2)(b2 + 2)(c2 + 2) ≥ 3(a + b + c)2. Because

(a + b + c)2 ≤ (|a| + |b| + |c|)2, we may assume that a, b, c are nonnegative. We willuse the fact that if x and y have the same sign then (1 + x)(1 + y) ≥ 1 + x + y. So,we write the inequality in the form∏(

a2 − 13

+ 1)≥ (a + b + c)2

9

and we have three cases

i) If a, b, c are at least 1, then∏(

a2 − 13

+ 1)≥ 1 +

∑ a2 − 13

≥

(∑a)2

9.

ii) If two of the three numbers are at least 1, let them be a and b, then we have

∏(a2 − 1

3+ 1)

≥(

1 +a2 − 1

3+

b2 − 13

)(c2 + 2

3

)=

(a2 + b2 + 1)9

(12 + 12 + c2)9

≥ (a + b + c)2

9by the Cauchy-Schwarz Inequality.iii) If all three numbers are at most 1, then by Bernoulli Inequality we have

∏(a2 − 1

3+ 1)≥ 1 +

∑ a2 − 13

≥

(∑a)2

9

and the proof is complete.

68 Solutions

Second solution:Expanding everything, we reduce the problem to proving that

(abc)2 + 2∑

a2b2 + 4∑

a2 + 8 ≥ 9∑

ab.

Because 3∑

a2 ≥ 3∑

ab and 2∑

a2b2 + 6 ≥ 4∑

ab, we are left with the

inequality (abc)2 +∑

a2 + 2 ≥ 2∑

ab. Of course, we can assume that a, b, c arenon-negative and we can write a = x2, b = y2, c = z2. In this case

2∑

ab−∑

a2 = (x + y + z)(x + y − z)(y + z − x)(z + x− y).

It is clear that if x, y, z are not side lengths of a triangle, then the inequalityis trivial. Otherwise, we can take x = u + v, y = v + w, z = w + u and reduce theinequality to

((u + v)(v + w)(w + u))4 + 2 ≥ 16(u + v + w)uvw.

We have ((u + v)(v + w)(w + u))4 + 1 + 1 ≥ 3 3√

(u + v)4(v + w)4(u + w)4 andit remains to prove that the last quantity is at least 16(u + v + w)uvw. This comesdown to

(u + v)4(v + w)4(w + u)4 ≥ 163

33(uvw)3(u + v + w)3.

But this follows from the known inequalities

(u + v)(v + w)(w + u) ≥ 89(u + v + w)(uv + vw + wu),

(uv + vw + wu)4 ≥ 34(uvw)83 , u + v + w ≥ 3 3

√uvw.

Third solution:In the same manner as in the Second solution, we reduce the problem to proving

that(abc)2 + 2 ≥ 2

∑ab−

∑a2.

Now, using Schur’s Inequality, we infer that

2∑

ab−∑

a2 ≤ 9abc

a + b + c

and as an immediate consequence of the AM-GM Inequality we have9abc

a + b + c≤ 3 3

√(abc)2.

This shows that as soon as we prove that

(abc)2 + 2 ≥ 3 3√

(abc)2,

the problem is solved. But the last assertion follows from the AM-GM Inequality.


68. [ Vasile Cırtoaje ] Prove that if 0 < x ≤ y ≤ z and x + y + z = xyz + 2, thena) (1− xy)(1− yz)(1− xz) ≥ 0;

b) x2y ≤ 1, x3y2 ≤ 3227

.

Solution:a) We have

(1− xy)(1− yz) = 1− xy− yz + xy2z = 1− xy− yz + y(x + y + z− 2) = (y− 1)2 ≥ 0

and similarly

(1− yz)(1− zx) = (1− z)2 ≥ 0, (1− zx)(1− xy) = (1− x)2 ≥ 0.

So the expressions 1− xy, 1− yz and 1− zx have the same sign.b) We rewrite the relation x+y+z = xyz+2 as (1−x)(1−y)+(1−z)(1−xy) = 0.

If x > 1 then z ≥ y ≥ x > 1 and so (1− x)(1− y) + (1− z)(1− xy) > 0, impossible.So we have x ≤ 1. Next we distinguish two cases 1) xy ≤ 1; 2) xy > 1.

1) xy ≤ 1. We have x2y ≤ x ≤ 1 and x3y2 ≤ x ≤ 1 <3227

.

2) xy > 1. From y ≥ √xy we get y > 1. Next we rewrite the relation x + y + z =xyz + 2 as x + y − 2 = (xy − 1)z. Because z ≥ y gives x + y − 2 ≥ (xy − 1)y,(y − 1)(2 − x − xy) ≥ 0 so 2 ≥ x(1 + y). Using the AM-GM Inequality, we have

1 + y ≥ 2√

y and 1 + y = 1 +y

2+

y

2≥ 3 3

√1 · y

2· y

2. Thus we have 2 ≥ 2x

√y and

2 ≥ 3x3

√y2

4, which means that x2y ≤ 1 and x3y2 ≤ 32

27.

The equality x2y = 1 takes place when x = y = 1 and the equality x3y2 =3227

takes place when x =23, y = z = 2.

69. [ Titu Andreescu ] Let a, b, c be positive real numbers such that a+b+c ≥ abc.

Prove that at least two of the inequalities

2a

+3b

+6c≥ 6,

2b

+3c

+6a≥ 6,

2c

+3a

+6b≥ 6,

are true.TST 2001, USA

Solution:The most natural idea is to male the substitution

1a

= x,1b

= y,1c

= z. Thus, wehave x, y, z > 0 and xy + yz + zx ≥ 1 and we have to prove that at least two of theinequalities 2x+3y+6z ≥ 6, 2y+3z+6x ≥ 6, 2z+3x+6y ≥ 6 are true. Suppose this isnot the case. Then we may assume that 2x+3y+6z < 6 and 2z+3x+6y < 6. Adding,

we find that 5x+9y+8z < 12. But we have x ≥ 1− yz

y + z. Thus, 12 >

5− 5yz

y + z+9y+8z

70 Solutions

which is the same as 12(y+z) > 5+9y2 +8z2 +12yz ⇔ (2z−1)2 +(3y+2z−2)2 < 0,which is clearly impossible. Thus, the conclusion follows.

70. [ Gabriel Dospinescu, Marian Tetiva ] Let x, y, z > 0 such that

x + y + z = xyz.

Prove that(x− 1) (y − 1) (z − 1) ≤ 6

√3− 10.

First solution:Because of x < xyz ⇒ yz > 1 (and the similar relations xz > 1, xy > 1) at most

one of the three numbers can be less than 1. In any of these cases (x ≤ 1, y ≥ 1,z ≥ 1 or the similar ones) the inequality to prove is clear. The only case we still haveto analyse is that when x ≥ 1, y ≥ 1 and z ≥ 1.

In this situation denote

x− 1 = a, y − 1 = b, z − 1 = c.

Then a, b, c are nonnegative real numbers and, because

x = a + 1, y = b + 1, z = c + 1,

they satisfya + 1 + b + 1 + c + 1 = (a + 1) (b + 1) (c + 1) ,

which meansabc + ab + ac + bc = 2.

Now let x = 3√

abc; we have

ab + ac + bc ≥ 3 3√

abacbc = 3x2,

that’s why we get

x3 + 3x2 ≤ 2 ⇔ (x + 1)(x2 + 2x− 2

)≤ 0 ⇔

⇔ (x + 1)(x + 1 +

√3) (

x + 1−√

3)≤ 0.

For x ≥ 0, this yields3√

abc = x ≤√

3− 1,

or, equivalently

abc ≤(√

3− 1)3

,

which is exactly(x− 1) (y − 1) (z − 1) ≤ 6

√3− 10.

The proof is complete.


Second solution:Like in the first solution (and due to the symmetry) we may suppose that x ≥ 1,

y ≥ 1; we can even assume that x > 1, y > 1 (for x = 1 the inequality is plain). Thenwe get xy > 1 and from the given condition we have

z =x + y

xy − 1.

The relation to prove is

(x− 1) (y − 1) (z − 1) ≤ 6√

3− 10 ⇔⇔ 2xyz − (xy + xz + yz) ≤ 6

√3− 9,

or, with this expression of z,

2xyx + y

xy − 1− xy − (x + y)

x + y

xy − 1≤ 6

√3− 9 ⇔

⇔ (xy − x− y)2 +(6√

3− 10)xy ≤ 6

√3− 9,

after some calculations.Now, we put x = a + 1, y = b + 1 and transform this into

a2b2 +(6√

3− 10)

(a + b + ab)− 2ab ≥ 0.

Buta + b ≥ 2

√ab

and 6√

3− 10 > 0, so it suffices to show that

a2b2 +(6√

3− 10)(

2√

ab + ab)− 2ab ≥ 0.

The substitution t =√

ab ≥ 0 reduces this inequality to

t4 +(6√

3− 12)

t2 + 2(6√

3− 10)

t ≥ 0,

ort3 +

(6√

3− 12)

t + 2(6√

3− 10)≥ 0.

The derivative of the function

f (t) = t3 +(6√

3− 12)

t + 2(6√

3− 10)

, t ≥ 0

is

f′(t) = 3

(t2 −

(√3− 1

)2)

and has only one positive zero. It is√

3−1 and it’s easy to see that this is a minimumpoint for f in the interval [0,∞). Consequently

f (t) ≥ f(√

3− 1)

= 0,

and we are done.A final observation: in fact we have

f (t) =(t−

√3 + 1

)2 (t + 2

√3− 2

),

72 Solutions

which shows that f (t) ≥ 0 for t ≥ 0.

71. [ Marian Tetiva ] Prove that for any positive real numbers a, b, c,∣∣∣∣a3 − b3

a + b+

b3 − c3

b + c+

c3 − a3

c + a

∣∣∣∣ ≤ (a− b)2 + (b− c)2 + (c− a)2

4.

Moldova TST, 2004

First solution:First of all, we observe that right hand side can be transformed into∑ a3 − b3

a + b= (a3 − b3)

(1

a + b− 1

a + c

)+ (b3 − c3)

(1

b + c− 1

a + c

)=

=(a− b)(c− b)(a− c)

(∑ab)

(a + b)(a + c)(b + c)

and so we have to prove the inequality

|(a− b)(b− c)(c− a)|(ab + bc + ca)(a + b)(b + c)(c + a)

≤ 12

(∑a2 −

∑ab)

.

It is also easy to prove that (a + b)(b + c)(c + a) ≥ 89(a + b + c)(ab + bc + ca) and

so we are left with29·∑

a ·(∑

(a− b)2)≥∣∣∣∏(a− b)

∣∣∣ .Using the AM-GM Inequality, we reduce this inequality to the following one

827

(∑a)3

≥∣∣∣∏(a− b)

∣∣∣ .This one is easy. Just observe that we can assume that a ≥ b ≥ c and in this case itbecomes

(a− b)(a− c)(b− c) ≤ 827

(a + b + c)3

and it follows from the AM-GM Inequality.

Second solution (by Marian Tetiva):It is easy to see that the inequality is not only cyclic, but symmetric. That is why

we may assume that a ≥ b ≥ c > 0. The idea is to use the inequality

x +y

2≥ x2 + xy + y2

x + y≥ y +

x

2,

which is true if x ≥ y > 0. The proof of this inequality is easy and we won’t insist.Now, because a ≥ b ≥ c > 0, we have the three inequalities

a +b

2≥ a2 + ab + b2

a + b≥ b +

a

2, b +

c

2≥ b2 + bc + c2

b + c≥ c +

b

2and of course

a +c

2≥ a2 + ac + c2

a + c≥ c +

a

2.


That is why we can write∑ a3 − b3

a + b= (a− b)

a2 + ab + b2

a + b+ (b− c)

b2 + bc + c2

b + c− (a− c)

a2 + ac + c2

a + c≥

≥ (a− b)(b +

a

2

)+ (b− c)

(c +

b

2

)− (a− c)

(a +

c

2

)=

= −

∑(a− b)2

4.

In the same manner we can prove that∑ a3 − b3

a + b≤∑

(a− b)2

4and the conclusion follows.

72. [ Titu Andreescu ] Let a, b, c be positive real numbers. Prove that

(a5 − a2 + 3)(b5 − b2 + 3)(c5 − c2 + 3) ≥ (a + b + c)3.

USAMO, 2004

Solution:We start with the inequality a5− a2 + 3 ≥ a3 + 2 ⇔ (a2− 1)(a3− 1) ≥ 0. Thus,

it remains to show that ∏(a3 + 2

)≥(∑

a)3

.

Using the AM-GM Inequality, one has

a3

a3 + 2+

1b3 + 2

+1

c3 + 2≥ 3a

3

√∏(a3 + 2)

.

We write two similar inequalities and then add up all these relations. We will findthat ∏

(a3 + 2) ≥(∑

a)3

,

which is what we wanted.

73. [ Gabriel Dospinescu ] Let n > 2 and x1, x2, . . . , xn > 0 such that(n∑

k=1

xk

)(n∑

k=1

1xk

)= n2 + 1.

Prove that (n∑

k=1

x2k

)·

(n∑

k=1

1x2

k

)> n2 + 4 +

2n(n− 1)

.

74 Solutions

Solution:In this problem, a combination between identities and the Cauchy-Schwarz

Inequality is the way to proceed. So, let us start with the expression∑1≤i<j≤n

(xi

xj+

xj

xi− 2)2

.

We can immediately see that∑1≤i<j≤n

(xi

xj+

xj

xi− 2)2

=∑

1≤i<j≤n

(x2

i

x2j

+x2

j

x2i

− 4 · xi

xj− 4 · xj

xi+ 6

)=

=

(n∑

i=1

x2i

)(n∑

i=1

1x2

i

)− 4

(n∑

i=1

xi

)(n∑

i=1

1xi

)+ 3n2.

Thus we could find from the inequality∑1≤i<j≤n

(xi

xj+

xj

xi− 2)2

≥ 0

that (n∑

i=1

x2i

)(n∑

i=1

1x2

i

)≥ n2 + 4.

Unfortunately, this is not enough. So, let us try to minimize∑1≤i<j≤n

(xi

xj+

xj

xi− 2)2

.

This could be done using the Cauchy-Schwarz Inequality:

∑1≤i<j≤n

(xi

xj+

xj

xi− 2)2

≥

∑1≤i<j≤n

(xi

xj+

xj

xi− 2)2

(n

2

) .

Because∑

i≤i<j≤n

(xi

xj+

xj

xi− 2)

= 1, we deduce that

(n∑

i=1

x2i

)(n∑

i=1

1x2

)≥ n2 + 4 +

2n(n− 1)

,

which is what we wanted to prove. Of course, we should prove that we cannot haveequality. But equality would imply that x1 = x2 = · · · = xn, which contradicts theassumption (

n∑k=1

xk

)(n∑

k=1

1xk

)= n2 + 1.


74. [ Gabriel Dospinescu, Mircea Lascu, Marian Tetiva ] Prove that for any po-sitive real numbers a, b, c,

a2 + b2 + c2 + 2abc + 3 ≥ (1 + a)(1 + b)(1 + c).

First solution:Let f(a, b, c) = a2 + b2 + c2 + abc + 2− a− b− c− ab− bc− ca. We have to prove

that all values of f are nonnegative. If a, b, c > 3, then we clearly have1a

+1b

+1c

< 1,

which means that f(a, b, c) > a2 + b2 + c2 + 2 − a − b − c > 0. So, we may assume

that a ≤ 3 and let m =b + c

2. Easy computations show that f(a, b, c)− f(a,m,m) =

(3− a)(b− c)2

4≥ 0 and so it remains to prove that f(a,m,m) ≥ 0, which is the same

as(a + 1)m2 − 2(a + 1)m + a2 − a + 2 ≥ 0.

This is clearly true, because the discriminant of the quadratic equation is −4(a +1)(a− 1)2 ≤ 0.

Second solution:Recall Turkevici’s Inequality

x4 + y4 + z4 + t4 + 2xyzt ≥ x2y2 + y2z2 + z2t2 + t2x2 + x2z2 + y2t2

for all positive real numbers x, y, z, t. Taking t = 1, a = x2, b = y2, x = z2 and usingthe fact that 2

√abc ≤ abc + 1, we find the desired inequality.

75. [ Titu Andreescu, Zuming Feng ] Let a, b, c be positive real numbers. Provethat

(2a + b + c)2

2a2 + (b + c)2+

(2b + a + c)2

2b2 + (a + c)2+

(2c + a + b)2

2c2 + (a + b)2≤ 8.

USAMO, 2003

First solution:Because the inequality is homogeneous, we can assume that a + b + c = 3. Then

(2a + b + c)2

2a2 + (b + c)2=

a2 + 6a + 93a2 − 6a + 9

=13

(1 + 2 · 4a + 3

2 + (a− 1)2)≤

≤ 13

(1 + 2 · 4a + 3

2

)=

4a + 43

.

Thus ∑ (2a + b + c)2

2a2 + (b + c)2≤ 1

3

∑(4a + 4) = 8.

76 Solutions

Second solution:

Denote x =b + c

a, y =

c + a

b, z =

a + b

c. We have to prove that

∑ (x + 2)2

x2 + 2≤ 8 ⇔

∑ 2x + 1x2 + 2

≤ 52⇔∑ (x− 1)2

x2 + 2≥ 1

2.

But, from the Cauchy-Schwarz Inequality, we have∑ (x− 1)2

x2 + 2≥ (x + y + z − 3)2

x2 + y2 + z2 + 6.

It remains to prove that

2(x2 + y2 + z2 + 2xy + 2yz + 2zx− 6x− 6y − 6z + 9) ≥ x2 + y2 + z2 + 6 ⇔

⇔ x2 + y2 + z2 + 4(xy + yz + zx)− 12(x + y + z) + 12 ≥ 0.

Now xy +yz + zx ≥ 3 3√

x2y2z2 ≥ 12 (because xyz ≥ 8), so we still have to prove that(x + y + z)2 + 24− 12(x + y + z) + 12 ≥ 0, which is equivalent to (x + y + z− 6)2 ≥ 0,clearly true.

76. Prove that for any positive real numbers x, y and any positive integers m,n,

(n−1)(m−1)(xm+n+ym+n)+(m+n−1)(xmyn+xnym) ≥ mn(xm+n−1y+ym+n−1x).

Austrian-Polish Competition, 1995

Solution:We transform the inequality as follows:

mn(x− y)(xm+n−1 − ym+n−1) ≥ (m + n− 1)(xm − ym)(xn − yn) ⇔

⇔ xm+n−1 − ym+n−1

(m + n− 1)(x− y)≥ xm − ym

m(x− y)· xn − yn

n(x− y)(we have assumed that x > y). The last relation can also be written

(x− y)∫ x

y

tm+n−2dt ≥∫ x

y

tm−1dt ·∫ x

y

tn−1dt

and this follows from Chebyshev’s Inequality for integrals.

77. Let a, b, c, d, e be positive real numbers such that abcde = 1. Prove that

a + abc

1 + ab + abcd+

b + bcd

1 + bc + bcde+

c + cde

1 + cd + cdea+

d + dea

1 + de + deab+

e + eab

1 + ea + eabc≥ 10

3.

Crux Mathematicorum

Solution:We consider the standard substitution

a =x

y, b =

y

z, c =

z

t, d =

t

u, e =

u

x


with x, y, z, t, u > 0. It is clear that

a + abc

1 + ab + abcd=

1y

+1t

1x

+1z

+1u

.

Writing the other relations as well, and denoting1x

= a1,1y

= a2,1z

= a3,1t

= a4,1u

=

a5, we have to prove that if ai > 0, then∑ a2 + a4

a1 + a3 + a5≥ 1

3.

Using the Cauchy-Schwarz Inequality, we minor the left-hand side with

4S2

2S2 − (a2 + a4)2 − (a1 + a4)2 − (a3 + a5)2 − (a2 + a5)2 − (a1 + a3)2,

where S =5∑

i=1

ai. By applying the Cauchy-Schwarz Inequality again for the

denominator of the fraction, we obtain the conclusion.

78. [ Titu Andreescu ] Prove that for any a, b, c, ∈(0,

π

2

)the following inequality

holdssin a · sin(a− b) · sin(a− c)

sin(b + c)+

sin b · sin(b− c) · sin(b− a)sin(c + a)

+sin c · sin(c− a) · sin(c− b)

sin(a + b)≥ 0.

TST 2003, USA

Solution:Let x = sin a, y = sin b, z = sin c. Then we have x, y, z > 0. It is easy to see that

the following relations are true:

sin a · sin(a− b) · sin(a− c) · sin(a + b) · sin(a + c) = x(x2 − y2)(x2 − z2)

Using similar relations for the other terms, we have to prove that:∑x(x2 − y2)(x2 − y2) ≥ 0.

With the substitution x =√

u, y =√

v, z =√

w the inequality becomes∑√

u(u−v)(u− w) ≥ 0. But this follows from Schur’s Inequality.

79. Prove that if a, b, c are positive real numbers then,√a4 + b4 + c4 +

√a2b2 + b2c2 + c2a2 ≥

√a3b + b3c + c3a +

√ab3 + bc3 + ca3.

KMO Summer Program Test, 2001

Solution:It is clear that it suffices to prove the following inequalities∑

a4 +∑

a2b2 ≥∑

a3b +∑

ab3

78 Solutions

and (∑a4)(∑

a2b2)≥(∑

a3b)(∑

ab3)

.

The first one follows from Schur’s Inequality∑a4 + abc

∑a ≥

∑a3b +

∑ab3

and the fact that ∑a2b2 ≥ abc

∑a.

The second one is a simple consequence of the Cauchy-Schwarz Inequality:

(a3b + b3c + c3a)2 ≤ (a2b2 + b2c2 + c2a2)(a4 + b4 + c4)

(ab3 + bc3 + ca3)2 ≤ (a2b2 + b2c2 + c2a2)(a4 + b4 + c4).

80. [ Gabriel Dospinescu, Mircea Lascu ] For a given n > 2 find the smallestconstant kn with the property: if a1, . . . , an > 0 have product 1, then

a1a2

(a21 + a2)(a2

2 + a1)+

a2a3

(a22 + a3)(a2

3 + a2)+ · · ·+ ana1

(a2n + a1)(a2

1 + an)≤ kn.

Solution:Let us take first a1 = a2 = · · · = an−1 = x, an =

1xn−1

. We infer that

kn ≥2x2n−1

(xn+1 + 1)(x2n−1 + 1)+

n− 2(1 + x)2

>n− 2

(1 + x)2

for all x > 0. Clearly, this implies kn ≥ n− 2. Let us prove that n − 2 is a goodconstant and the problem will be solved.

First, we will prove that (x2 + y)(y2 + x) ≥ xy(1 + x)(1 + y). Indeed, this is thesame as (x + y)(x− y)2 ≥ 0. So, it suffices to prove that

1(1 + a1)(1 + a2)

+1

(1 + a2)(1 + a3)+ · · ·+ 1

(1 + an)(1 + a1)≤ n− 2.

Now, we take a1 =x1

x2, . . . , an =

xn

x1and the above inequality becomes

n∑k=1

(1− xk+1xk+2

(xk + xk+1)(xk+1 + xk+2)

)≥ 2,

which can be also written in the following fromn∑

k=1

xk

xk + xk+1+

x2k+1

(xk + xk+1)(xk+1 + xk+2)≥ 2.

Clearly,n∑

k=1

xk

xk + xk+1≥

n∑k=1

xk

x1 + x2 + · · ·+ xn= 1.


So, we have to prove the inequalityn∑

k=1

x2k+1

(xk + xk+1)(xk+1 + xk+2)≥ 1.

Using the Cauchy-Schwarz Inequality, we infer that

n∑k=1

x2k+1

(xk + xk+1)(xk+1 + xk+2)≥

(n∑

k+1

xk

)2

n∑k=1

(xk + xk+1)(xk+1 + xk+2)

and so it suffices to prove that(n∑

k=1

xk

)2

≥n∑

k=1

x2k + 2

n∑k=1

xkxk+1 +n∑

k=1

xkxk+2.

But this one is equivalent to

2∑

1≤i<j≤n

xixj ≥ 2n∑

k=1

xkxk+1 +n∑

k=1

xkxk+2

and it is clear. Thus, kn = n− 2.

81. [ Vasile Cırtoaje ] For any real numbers a, b, c, x, y, z prove that the inequalityholds

ax + by + cz +√

(a2 + b2 + c2)(x2 + y2 + z2) ≥ 23(a + b + c)(x + y + z).

Kvant, 1989

Solution:

Let us denote t =

√x2 + y2 + z2

a2 + b2 + c2. Using the substitutions x = tp, y = tq and

z = tr, which implya2 + b2 + c2 = p2 + q2 + r2.

The given inequality becomes

ap + bq + cr + a2 + b2 + c2 ≥ 23(a + b + c)(p + q + r),

(a + p)2 + (b + q)2 + (c + r)2 ≥ 43(a + b + c)(p + q + r).

Since4(a + b + c)(p + q + r) ≤ [(a + b + c) + (p + q + r)]2,

it suffices to prove that

(a + p)2 + (b + q)2 + (c + r)2 ≥ 13[(a + p) + (b + q) + (c + r)]2.

This inequality is clearly true.

80 Solutions

82. [ Vasile Cırtoaje ] Prove that the sides a, b, c of a triangle satisfy the inequality

3(

a

b+

b

c+

c

a− 1)≥ 2

(b

a+

c

b+

a

c

).

First solution:We may assume that c is the smallest among a, b, c. Then let x = b− a + c

2. After

some computations, the inequality becomes

(3a−2c)x2+(x + c− a

4

)(a−c)2 ≥ 0 ⇔ (3a−2c)(2b−a−c)2+(4b+2c−3a)(a−c)2 ≥ 0

which follows immediately from 3a ≥ 2c, 4b + 2c− 3a = 3(b + c− a) + b− c > 0.

Second solution:Make the classical substitution a = y + z, b = z + x, c = x + y and clear denomi-

nators. The problem reduces to proving that

x3 + y3 + z3 + 2(x2y + y2z + z2x) ≥ 3(xy2 + yz2 + zx2).

We can of course assume that x is the smallest among x, y, z. Then we can writey = x + m, z = x + n with nonnegatives m and n. A short computation shows thatthe inequality reduces to 2x(m2 −mn + n2) + m3 + n3 + 2m2n − 3n2m ≥ 0. All weneed to prove is that m3 + n3 + 2m2n ≥ 3n2m ⇔ (n−m)3 − (n−m)m2 + m3 ≥ 0and this follows immediately from the inequality t3 + 1 ≥ 3, true for t ≥ −1.

83. [ Walther Janous ] Let n > 2 and let x1, x2, . . . , xn > 0 add up to 1. Provethat

n∏i=1

(1 +

1xi

)≥

n∏i=1

(n− xi

1− xi

).

Crux Mathematicorum

First solution:The most natural idea is to use the fact that

n− xi

1− xi= 1 +

n− 1x1 + x2 + · · ·+ xi−1 + xi+1 + · · ·+ xn

.

Thus, we haven∏

i=1

(n− xi

1− xi

)≤

n∏i=1

(1 +

1n−1√

x1x2 . . . xi−1xi+1 . . . xn

)and we have to prove the inequality

n∏i=1

(1 +

1xi

)≥

n∏i=1

(1 +

1n−1√

x1x2 . . . xi−1xi+1 . . . xn

).


But this one is not very hard, because it follows immediately by multiplying theinequalities

∏j 6=i

(1 +

1xj

)≥

1 + n−1

√∏j 6=i

1xj

n−1

obtained from Huygens Inequality.

Second solution:We will prove even more, that

n∏i=1

(1 +

1xi

)≥(

n2 − 1n

)n

·n∏

i=1

11− xi

.

It is clear that this inequality is stronger than the initial one. First, let us prove that

n∏i=1

1 + xi

1− xi≥(

n + 1n− 1

)n

.

This follows from Jensen’s Inequality for the convex function f(x) = ln(1 + x) −ln(1− x). So, it suffices to prove that(

n + 1n− 1

)n

n∏i=1

xi

·n∏

i=1

(1− xi)2 ≥(

n2 − 1n

)n

.

But a quick look shows that this is exactly the inequality proved in the solution ofthe problem 121.

84. [ Vasile Cırtoaje, Gheorghe Eckstein ] Consider positive real numbersx1, x2, ..., xn such that x1x2...xn = 1. Prove that

1n− 1 + x1

+1

n− 1 + x2+ ... +

1n− 1 + xn

≤ 1.

TST 1999, Romania

First solution:Suppose the inequality is false for a certain system of n numbers. Then we can

find a number k > 1 and n numbers which add up to 1, let them be ai, such that1

n− 1 + xi= kai. Then we have

1 =n∏

i=1

(1

kai− n + 1

)<

n∏i=1

(1ai− n + 1

).

82 Solutions

We have used here the fact that ai <1

n− 1. Now, we write 1 − (n − 1)ak = bk and

we find thatn∑

k=1

bk = 1 and alson∏

k=1

(1− bk) < (n− 1)nb1 . . . bn. But this contradicts

the fact that for each j we have

1− bj = b1 + · · ·+ bj−1 + bj+1 + · · ·+ bn ≥ (n− 1) n−1√

b1 . . . bj−1bj+1 . . . bn.

Second solution:Let us write the inequality in the form

x1

n− 1 + x1+

x2

n− 1 + x2+ . . . +

xn

n− 1 + xn≥ 1.

This inequality follows by summing the following inequalities

x1

n− 1 + x1≥ x

1− 1n

1

x1− 1

n1 + x

1− 1n

2 + . . . + x1− 1

nn

, . . . ,xn

n− 1 + xn≥ x

1− 1n

n

x1− 1

n1 + x

1− 1n

2 + . . . + x1− 1

nn

,

The first from these inequalities is equivalent to

x1− 1

n1 + x

1− 1n

2 + . . . + x1− 1

nn ≥ (n− 1)x−

1n

1

and follows from the AM-GM Inequality.

Remark.Replacing the numbers x1, x2, . . . , xn with

1x1

,1x2

, . . . ,1xn

respectively, the in-

equality becomes as follows1

1 + (n− 1)x1+

11 + (n− 1)x2

+ . . . +1

1 + (n− 1)xn≥ 1.

85. [ Titu Andreescu ] Prove that for any nonnegative real numbers a, b, c suchthat a2 + b2 + c2 + abc = 4 we have 0 ≤ ab + bc + ca− abc ≤ 2.

USAMO, 2001

First solution (by Richard Stong):The lower bound is not difficult. Indeed, we have ab + bc + ca ≥ 3 3

√a2b2c2 and

thus it is enough to prove that abc ≤ 3 3√

a2b2c2, which follows from the fact thatabc ≤ 4. The upper bound instead is hard. Let us first observe that there are twonumbers among the three ones, which are both greater or equal than 1 or smallerthan or equal to 1. Let them be b and c. Then we have

4 ≥ 2bc + a2 + abc ⇒ (2− a)(2 + a) ≥ bc(2 + a) ⇒ bc ≤ 2− a.

Thus, ab+bc+ca−abc ≤ ab+2−a+ac−abc and it is enough to prove the inequalityab + 2− a + ac− abc ≤ 2 ⇔ b + c− bc ≤ 1 ⇔ (b− 1)(c− 1) ≥ 0, which is true due toour choice.


Second solution:We won’t prove again the lower part, since this is an easy problem. Let us con-

centrate on the upper bound. Let a ≥ b ≥ c and let a = x + y, b = x − y. Thehypothesis becomes x2(2 + c) + y2(2 − c) = 4 − c2 and we have to prove that

(x2 − y2)(1 − c) ≤ 2(1 − xc). Since y2 = 2 + c − 2 + c

2− cx2, the problem asks to

prove the inequality4x2 − (4− c2)

2− c(1 − c) ≤ 2(1 − cx). Of course, we have c ≤ 1

and 0 ≤ y2 = 2 + c − 2+c2−cx2 ⇒ x2 ≤ 2 − c ⇒ x ≤

√2− c (we have used

the fact that a ≥ b ⇒ y ≥ 0 and b ≥ 0 ⇒ x ≥ y ≥ 0). Now, consider the

function f : [0,√

2− c] → R, f(x) = 2(1 − cx) − 4x2 − (4− c2)2− c

(1 − c). We have

f ′(x) = −2c − 8x · 1− c

2− c≤ 0 and thus f is decreasing and f(x) ≥ f(

√2− c). So, we

have to prove that

f(√

2− c) ≥ 0 ⇔ 2(1−c√

2− c) ≥ (2−c)(1−c) ⇔ 3 ≥ c+2√

2− c ⇔ (1−√

2− c)2 ≥ 0

clearly true. Thus, the problem is solved.

86. [ Titu Andreescu ] Prove that for any positive real numbers a, b, c the followinginequality holds

a + b + c

3− 3√

abc ≤ max{(√

a−√

b)2, (√

b−√

c)2, (√

c−√

a)2}.

TST 2000, USA

Solution:A natural idea would be to assume the contrary, which means that

a + b + c

3− 3√

abc ≤ a + b− 2√

ab

a + b + c

3− 3√

abc ≤ b + c− 2√

bc

a + b + c

3− 3√

abc ≤ c + a− 2√

ca.

Adding these inequalities, we find that

a + b + c− 3 3√

abc > 2(a + b + c−√

ab−√

bc−√

ca).

Now, we will prove that a + b + c− 3 3√

abc ≤ 2(a + b + c−√

ab−√

bc−√

ca) and theproblem will be solved. Since the above inequality is homogeneous, we may assumethat abc = 1. Then, it becomes 2

√ab + 2

√bc + 2

√ca − a − b − c ≤ 3. Now, using

Schur’s inequality, we find that for any positive reals x,y,z we have:

2xy + 2yz + 2zx− x2 − y2 − z2 ≤ 9xyz

x + y + z≤ 3 3

√x2y2z2

All we have to do is take x =√

a, y =√

b, z =√

c in the above inequality.

84 Solutions

87. [ Kiran Kedlaya ] Let a,b,c be positive real numbers. Prove that

a +√

ab + 3√

abc

3≤ 3

√a · a + b

2· a + b + c

3.

Solution (by Anh Cuong):We have that

a +√

ab + 3√

abc ≤ a + 3

√ab

a + b

2+ 3√

abc.

Now we will prove that

a + 3

√ab

a + b

2+ 3√

abc ≤ 3

√a · a + b

2· a + b + c

3.

By the AM-GM Inequality, we have

3

√1 · 2a

a + b· 3a

a + b + c≤

1 +2a

a + b+

3a

a + b + c3

,

3

√1 · 1 · 3b

a + b + c≤

2 +3b

a + b + c3

,

3

√1 · 2b

a + b· 3c

a + b + c≤

1 +2b

a + b+

3c

a + b + c3

.

Now, just add them up and we have the desired inequality. The equality occurs whena = b = c.

88. Find the greatest constant k such that for any positive integer n which is nota square, |(1 +

√n) sin(π

√n)| > k.

Vietnamese IMO Training Camp, 1995

Solution:We will prove that

π

2is the best constant. We must clearly have k <(

1 +√

i2 + 1)| sin

(π√

i2 + 1)| for all positive integers i. Because | sin

(π√

i2 + 1)| =

sinπ

i +√

i2 + 1, we deduce that

π

i +√

i2 + 1≥ sin

π

i +√

i2 + 1>

k

1 +√

i2 + 1, from

where it follows that k ≤ π

2. Now, let us prove that this constant is good. Clearly, the

inequality can be written

sin(π · {

√n})

>π

2(1 +√

n).

We have two cases


i) The first case is when {√

n} ≤ 12. Of course,

{√

n} ≥√

n−√

n− 1 =1

√n +

√n− 1

,

and because sin x ≥ x− x3

6we find that

sin(π{√

n}) ≥ sinπ√

n− 1 +√

n≥(

π√n− 1 +

√n

)− 1

6

(π√

n− 1 +√

n

)3

.

Let us prove that the last quantity is at leastπ

2(1 +√

n). This comes down to

2 +√

n−√

n− 11 +

√n

>π2

3(√

n +√

n− 1)2 ,

or 6(√

n +√

n− 1)2 + 3(√

n +√

n− 1) > π2(1 +√

n) and it is clear.

ii) The second case is when {√

n} >12. Let x = 1 − {

√n} <

12

and let n =

k2 + p, 1 ≤ p ≤ 2k. Because {√

n} >12⇒ p ≥ k + 1. Then it is easy to see that

x ≥ 1k + 1 +

√k2 + 2k

and so it suffices to prove that

sinπ

k + 1 +√

k2 + 2k≥ π

2(1 +√

k2 + k).

Using again the inequality sinx ≥ x− x3

6we infer that,

2√

k2 + k −√

k2 + 2k − k + 11 +

√k2 + k

>π2

3(1 + k +

√k2 + 2k

)2 .

But from the Cauchy-Schwarz Inequality we have 2√

k2 + k−√

k2 + 2k− k ≥ 0.

Because the inequality(1 + k +

√k2 + 2k

)2>

π2

3

(1 +

√k2 + k

)holds, this case is

also solved.

89. [ Dung Tran Nam ] Let x, y, z > 0 such that (x + y + z)3 = 32xyz. Find the

minimum and maximum ofx4 + y4 + z4

(x + y + z)4.

Vietnam, 2004First solution (by Tran Nam Dung):

We may of course assume that x + y + z = 4 and xyz = 2. Thus, we have to find the

86 Solutions

extremal values ofx4 + y4 + z4

44. Now, we have

x4 + y4 + z4 = (x2 + y2 + z2)2 − 2∑

x2y2 =

= (16− 2∑

xy)2 − 2(∑

xy)2 + 4xyz(x + y + z) =

= 2a2 − 64a + 288,

where a = xy+yz+zx. Because y+z = 4−x and yz =2x

, we must have (4−x)2 ≥ 8x

,

which implies that 3 −√

5 ≤ x ≤ 2. Due to symmetry, we have x, y, z ∈ [3 −√

5, 2].This means that (x− 2)(y − 2)(z − 2) ≤ 0 and also

(x− 3 +√

5)(y − 3 +√

5)(z − 3 +√

5) ≥ 0.

Clearing paranthesis, we deduce that

a ∈

[5,

5√

5− 12

].

But becausex4 + y4 + z4

44=

(a− 16)2 − 112128

, we find that the extremal values of the

expression are383− 165

√5

256,

9128

, attained for the triples

(3−

√5,

1 +√

52

,1 +

√5

2

),

respectively (2, 1, 1).

Second solution:As in the above solution, we must find the extremal values of x2 + y2 + z2 when

x + y + z = 1, xyz =132

, because after that the extremal values of the expression

x4 + y4 + z4 can be immediately found. Let us make the substitution x =a

4, y =

b

4, z =

c

2, where abc = 1, a + b + 2c = 4. Then x2 + y2 + z2 =

a2 + b2 + 4c2

16and so we

must find the extremal values of a2 +b2 +4c2. Now, a2 +b2 +4c2 = (4−2c)2− 2c

+4c2

and the problem reduces to finding the maximum and minimum of 4c2−8c− 1c

wherethere are positive numbers a, b, c such that abc = 1, a + b + 2c = 4. Of course, this

comes down to (4 − 2c)2 ≥ 4c, or to c ∈

[3−

√5

2, 1

]. But this reduces to the study

of the function f(x) = 4x2 − 8x− 1x

defined for

[3−

√5

2, 1

], which is an easy task.

90. [ George Tsintifas ] Prove that for any a, b, c, d > 0,

(a + b)3(b + c)3(c + d)3(d + a)3 ≥ 16a2b2c2d2(a + b + c + d)4.

Crux Mathematicorum


Solution:Let us apply Mac-Laurin Inequality for

x = abc, y = bcd, z = cda, t = dab.

We will find that∑ abc

4

3

≥

∑abc · bcd · cda

4=

a2b2c2d2∑

a

4.

Thus, it is enough to prove the stronger inequality

(a + b)(b + c)(c + d)(d + a) ≥ (a + b + c + d)(abc + bcd + cda + dab).

Now, let us observe that

(a + b)(b + c)(c + d)(d + a) = (ac + bd + ad + bc)(ac + bd + ab + cd) == (ac + bd)2 +

∑a2(bc + bd + cd) ≥ 4abcd +

∑a2(bc + bd + cd) =

= (a + b + c + d)(abc + bcd + cda + dab).

And so the problem is solved.

91. [ Titu Andreescu, Gabriel Dospinescu ] Find the maximum value of the ex-pression

(ab)n

1− ab+

(bc)n

1− bc+

(ca)n

1− cawhere a, b, c are nonnegative real numbers which add up to 1 and n is some positiveinteger.

Solution:First, we will treat the case n > 1. We will prove that the maximum value is1

3 · 4n−1. It is clear that ab, bc, ca ≤ 1

4and so

(ab)n

1− ab+

(bc)n

1− bc+

(ca)n

1− ca≤ 4

3((ab)n + (bc)n + (ca)n) .

Thus, we have to prove that (ab)n + (bc)n + (ca)n ≤ 14n

. Let a the maximum amonga, b, c. Then we have

14n≥ an(1− a)n = an(b + c)n ≥ anbn + ancn + nanbn−1c ≥ anbn + bncn + cnan.

So, we have proved that in this case the maximum is at most1

3 · 4n−1. But for

a = b =12, c = 0 this value is attained and this shows that the maximum value is

13 · 4n−1

for n > 1. Now, suppose that n = 1. In this case we have∑ ab

1− ab=∑ 1

1− ab− 3.

88 Solutions

Using the fact that a + b + c = 1, it is not difficult to prove that

11− ab

+1

1− bc+

11− ca

=3− 2

∑ab + abc

1−∑

ab + abc− a2b2c2.

We will prove that∑ ab

1− ab≤ 3

8. With the above observations, this reduces to

∑ab ≤ 3− 27(abc)2 + 19abc

11.

But using Schur’s Inequality we infer that∑ab ≤ 1 + 9abc

4and so it is enough to show that

9abc + 14

≤ 3− 27(abc)2 + 19abc

11⇔ 108(abc)2 + 23abc ≤ 1,

which is true because abc ≤ 127

.

Hence for n = 1, the maximum value is38

attained for a = b = c.

92. Let a, b, c be positive real numbers. Prove that1

a(1 + b)+

1b(1 + c)

+1

c(1 + a)≥ 3

3√

abc(1 + 3√

abc).

Solution:The following observation is crucial

(1 + abc)(

1a(1 + b)

+1

b(1 + c)+

1c(1 + a)

)+ 3 =

∑ 1 + abc + a + ab

a(1 + b)=

=∑ 1 + a

a(1 + b)+∑ b(c + 1)

1 + b.

We use now twice the AM-GM Inequality to find that∑ 1 + a

a(1 + b)+∑ b(c + 1)

1 + b≥ 3

3√

abc+ 3 3

√abc.

And so we are left with the inequality3

3√

abc+ 3 3

√abc− 3

1 + abc≥ 3

3√

abc(1 + 3√

abc),

which is in fact an identity!

93. [ Dung Tran Nam ] Prove that for any real numbers a, b, c such that a2 + b2 +c2 = 9,

2(a + b + c)− abc ≤ 10.

Vietnam, 2002


First solution (by Gheorghe Eckstein):Because max{a, b, c} ≤ 3 and |abc| ≤ 10, it is enough to consider only the cases

when a, b, c ≥ 0 or exactly of the three numbers is negative. First, we will supposethat a, b, c are nonnegative. If abc ≥ 1, then we are done, because

2(a + b + c)− abc ≤ 2√

3(a2 + b2 + c2)− 1 < 10.

Otherwise, we may assume that a < 1. In this case we have

2(a + b + c)− abc ≤ 2

(a + 2

√b2 + c2

2

)= 2a + 2

√18− 2a2 ≤ 10.

Now, assume that not all three numbers are nonnegative and let c < 0.Thus, the problem reduces to proving that for any nonnegative x, y, z whose

sum of squares is 9 we have 4(x + y − z) + 2xyz ≤ 20. But we can write this as(x−2)2+(y−2)2+(z−1)2 ≥ 2xyz − 6z − 2. Because 2xyz−6z−2 ≤ x(y2+z2)−6z−2 =−z3 + 3z − 2 = −(z − 1)2(z + 2) ≤ 0, the inequality follows.

Second solution:Of course, we have |a|, |b|, |c| ≤ 3 and |a + b + c|, |abc| ≤ 3

√3. Also, we may

assume of course that a, b, c are non-zero and that a ≤ b ≤ c. If c < 0 then we have2(a+b+c)−abc < −abc ≤ 3

√3 < 10. Also, if a ≤ b < 0 < c then we have 2(a+b+c) <

2c ≤ 6 < 10 + abc because abc > 0. If a < 0 < b ≤ c, using the Cauchy-SchwarzInequality we find that 2b+2c−a ≤ 9. Thus, 2(a+b+c) = 2b+2c−a+3a ≤ 9+3a

and it remains to prove that 3a − 1 ≤ abc. But a < 0 and 2bc ≤ 9 − a2, so that it

remains to show that9a− a3

2≥ 3a− 1 ⇔ (a + 1)2(a− 2) ≤ 0, which follows. So, we

just have to threat the case 0 < a ≤ b ≤ c. In this case we have 2b + 2c + a ≤ 9 and2(a + b + c) ≤ 9 + a. So, we need to prove that a ≤ 1 + abc. This is clear if a < 1 andif a > 1 we have b, c > 1 and the inequality is again. Thus, the problem is solved.

94. [ Vasile Cırtoaje ] Let a, b, c be positive real numbers. Prove that(a +

1b− 1)(

b +1c− 1)

+(

b +1c− 1)(

c +1a− 1)

+(

c +1a− 1)(

a +1b− 1)≥ 3.

First solution:With the notations x = a +

1b− 1, y = b +

1c− 1, z = c +

1a− 1, the inequality

becomes

xy + yz + zx ≥ 3.

We consider without loss of generality that x = max{x, y, z}. From

(x+1)(y+1)(z+1) = abc+1

abc+a+b+c+

1a+

1b+

1c≥ 2+a+b+c+

1a+

1b+

1c

= 5+x+y+z,

90 Solutions

we get

xyz + xy + yz + zx ≥ 4,

with equality if and only if abc = 1. Because y+z =1a

+b+(c− 1)2

c> 0 we distinguish

two cases a) x> 0, yz ≤ 0; b)x> 0,y> 0,z> 0.a) x> 0, yz ≤ 0. We have xyz ≤ 0 and from xyz + xy + yz + zx ≥ 4 we get

xy + yz + zx ≥ 4 > 3.b) x> 0,y> 0,z> 0. We denote xy + yz + zx = 3d2 with d > 0. From the mean

inequalities we have

xy + yz + zx ≥ 3 3√

x2y2z2,

from which we deduce that xyz ≤ d3. On the basis of this result, from the inequalityxyz + xy + yz + zx ≥ 4, we obtain d3 + 3d2 ≥ 4, (d − 1)(d + 2)2 ≥ 0, d ≥ 1 soxy + yz + zx ≥ 3. With this the given inequality is proved. We have equality in thecase a = b = c = 1.

Second solution:Let u = x + 1, v = y + 1, w = z + 1. Then we have

uvw = u + v + w + abc +1

abc≥ u + v + w + 2.

Now, consider the function f(t) = 2t3+t2(u+v+w)−uvw. Because limt→∞

f(t) = ∞ and

f(1) ≤ 0, we can find a real number r ≥ 1 such that f(r) = 0. Consider the numbersm =

u

r, n =

v

r, p =

w

r. They verify mnp = m + n + p + 2 we deduce from problem 49

that mn + np + pm ≥ 2(m + n + p) ⇒ uv + vw + wu ≥ 2r(u + v + w) ≥ 2(u + v + w).But because u = x + 1, v = y + 1, w = z + 1, this last relation is equivalent toxy + yz + zx ≥ 3, which is what we wanted.

95. [ Gabriel Dospinescu ] Let n be an integer greater than 2. Find the greatestreal number mn and the least real number Mn such that for any positive real numbersx1, x2, . . . , xn (with xn = x0, xn+1 = x1),

mn ≤n∑

i=1

xi

xi−1 + 2(n− 1)xi + xi+1≤ Mn.

Solution:We will prove that mn =

12(n− 1)

, Mn =12. First, let us see that the inequality

n∑i=1

xi

xi−1 + 2(n− 1)xi + xi+1≥ 1

2(n− 1)


is trivial, because xi−1 + 2(n − 1)xi + xi+1 ≤ 2(n − 1) ·n∑

k=1

xk for all i. This shows

that mn ≥1

2(n− 1). Taking xi = xi, the expression becomes

1x + xn−1 + 2(n− 1)

+(n− 2)x

1 + 2(n− 1)x + x2+

xn−1

1 + 2(n− 1)xn−1 + xn−2

and taking the limit when x approaches 0, we find that mn ≤ 12(n− 1)

and thus

mn =1

2(n− 1).

Now, we will prove that Mn ≥ 12. Of course, it suffices to prove that for any

x1, x2, . . . , xn > 0 we haven∑

i=1

xi

xi−1 + 2(n− 1)xi + xi+1≤ 1

2.

But it is clear thatn∑

i=1

2xi

xi−1 + 2(n− 1)xi + xi+1≤

n∑i=1

2xi

2√

xi−1 · xi+1 + 2(n− 1)xi=

=n∑

i=1

1

n− 1 +√

xi−1 · xi+1

xi

.

Taking√

xi−1 · xi+1

xi= ai, we have to prove that if

n∏i=1

ai = 1 then

n∑i=1

1n− 1 + ai

≤ 1. But this has already been proved in the problem 84. Thus,

Mn ≥ 12

and because for x1 = x2 = · · · = xn we have equality, we deduce that

Mn =12, which solves the problem.

96. [ Vasile Cırtoaje ] If x, y, z are positive real numbers, then

1x2 + xy + y2

+1

y2 + yz + z2+

1z2 + zx + z2

≥ 9(x + y + z)2

.

Gazeta Matematica

Solution:Considering the relation

x2 + xy + y2 = (x + y + z)2 − (xy + yz + zx)− (x + y + z)z,

92 Solutions

we get(x + y + z)2

x2 + xy + y2=

1

1− xy + yz + zx

(x + y + z)2− z

x + y + z

,

or(x + y + z)2

x2 + xy + y2=

11− (ab + bc + ca)− c

,

where a =x

x + y + z, b =

y

x + y + z, c =

z

x + y + z. The inequality can be rewritten

as1

1− d− c+

11− d− b

+1

1− d− a≥ 9,

where a, b, c are positive reals with a + b + c = 1 and d = ab + bc + ca. After makingsome computations the inequality becomes

9d3 − 6d2 − 3d + 1 + 9abc ≥ 0

ord(3d− 1)2 + (1− 4d + 9abc) ≥ 0.

which is Schur’s Inequality.

97. [ Vasile Cırtoaje ] For any a, b, c, d > 0 prove that

2(a3 + 1)(b3 + 1)(c3 + 1)(d3 + 1) ≥ (1 + abcd)(1 + a2)(1 + b2)(1 + c2)(1 + d2).

Gazeta Matematica

Solution:Using Huygens Inequality∏

(1 + a4) ≥ (1 + abcd)4,

we notice that it is enough to show that that

24∏

(a3 + 1)4 ≥∏

(1 + a4)(1 + a2)4.

Of course, it suffices to prove that 2(a3 + 1)4 ≥ (a4 + 1)(a2 + 1)4 for any positivereal a. But (a2+1)4 ≤ (a+1)2(a3+1)2 and we are left with the inequality 2(a3+1)2 ≥(a + 1)2(a4 + 1) ⇔ 2(a2 − a + 1)2 ≥ a4 + 1 ⇔ (a− 1)4 ≥ 0, which follows.

98. Prove that for any real numbers a, b, c,

(a + b)4 + (b + c)4 + (c + a)4 ≥ 47(a4 + b4 + c4).

Vietnam TST, 1996


Solution:Let us make the substitution a + b = 2z, b + c = 2x, c + a = 2y. The inequality

becomes∑

(y + z − x)4 ≤ 28∑

x4. Now, we have the following chain of identities∑(y + z − x)4 =

∑(∑x2 + 2yz − 2xy − 2xz

)2

= 3(∑

x2)2

+ 4(∑

x2)

(∑(yz − xy − xz)

)+ 4

∑(xy + xz − yz)2 = 3

(∑x2)2

− 4(∑

xy)(∑

x2)

+

+16∑

x2y2 − 4(∑

xy)2

= 4(∑

x2)2

+ 16∑

x2y2 −(∑

x)4

≤ 28∑

x4

because(∑

x2)2

≤ 3∑

x4,∑

x2y2 ≤∑

x4.

99. Prove that if a, b, c are positive real numbers such that abc = 1, then

11 + a + b

+1

1 + b + c+

11 + c + a

≤ 12 + a

+1

2 + b+

12 + c

.

Bulgaria, 1997

Solution:Let x = a + b + c and y = ab + bc + ca. Using brute-force, it is easy to see that

the left hand side isx2 + 4x + y + 3x2 + 2x + y + xy

, while the right hand side is12 + 4x + y

9 + 4x + 2y. Now,

the inequality becomes

x2 + 4x + y + 3x2 + 2x + y + xy

− 1 ≤ 12 + 4x + y

9 + 4x + 2y− 1 ⇔ 2x + 3− xy

x2 + 2x + y + xy≤ 3− y

9 + 4x + 2y.

For the last inequality, we clear denominators. Then using the inequalities x ≥ 3, y ≥3, x2 ≥ 3y, we have

53x2y ≥ 5x2,

x2y

3≥ y2, xy2 ≥ 9x, 5xy ≥ 15x, xy ≥ 3y and x2y ≥ 27.

Summing up these inequalities, the desired inequality follows.

100. [ Dung Tran Nam ] Find the minimum value of the expression1a

+2b

+3c

where a, b, c are positive real numbers such that 21ab + 2bc + 8ca ≤ 12.Vietnam, 2001

First solution (by Dung Tran Nam):

Let1a

= x,2b

= y,3c

= z. Then it is easy to check that the condition of the problembecomes 2xyz ≥ 2x + 4y + 7z. And we need to minimize x + y + z. But

z(2xy − 7) ≥ 2x + 4y ⇒

2xy > 7

z ≥ 2x + 4y

2xy − 7

94 Solutions

Now, we transform the expression so that after one application of the AM-GM

Inequality the numerator 3xy − 7 should vanish x + y + z ≥ x + y +2x + 4y

2xy − 7=

x +112x

+ y − 72x

+2x +

14x

2xy − 7≥ x +

112x

+ 2

√1 +

7x2

. But, it is immediate to prove

that 2

√1 +

7x2≥

3 +7x

2and so x + y + z ≥ 3

2+ x +

9x≥ 15

2. We have equality for

x = 3, y =52, z = 2. Therefore, in the initial problem the answer is

152

, achieved for

a =13, b =

45, c =

32.

Second solution:We use the same substitution and reduce the problem to finding the minimum

value of x + y + z when 2xyz ≥ 2x + 4y + 7z. Applying the weighted the AM-GMInequality we find that

x + y + z ≥(

5x

2

) 25

(3y)13

(15z

4

) 415

.

And also 2x + 4y + 7z ≥ 1015 · 12

13 · 5 7

15 · x 15 · y 1

3 · z 715 . This means that (x+

+y + z)2(2x + 4y + 7z) ≥ 2252

xyz. Because 2xyz ≥ 2x + 4y + 7z, we will have

(x + y + z)2 ≥ 2254⇒ x + y + z ≥ 15

2

with equality for x = 3, y =52, z = 2.

101. [ Titu Andreescu, Gabriel Dospinescu ] Prove that for any x, y, z, a, b, c > 0such that xy + yz + zx = 3,

a

b + c(y + z) +

b

c + a(z + x) +

c

a + b(x + y) ≥ 3.

Solution:We will prove the inequality

a

b + c(y + z) +

b

c + a(z + x) +

c

a + b(x + y) ≥

√3(xy + yz + zx)

for any a, b, c, x, y, z. Because the inequality is homogeneous in x, y, z we can assumethat x+y + z = 1. But then we can apply the Cauchy-Schwarz Inequality so that


to obtaina

b + cx +

b

c + ay +

c

a + bz +

√3(xy + yz + zx) ≤

≤

√∑(a

b + c

)2√∑x2 +

√34

√∑xy +

√34

√∑xy ≤

≤

√∑(a

b + c

)2

+32

√∑x2 + 2

∑xy =

√∑(a

b + c

)2

+32.

Thus, we are left with the inequality

√∑(a

b + c

)2

+32≤∑ a

b + c. But this one

is equivalent to∑ ab

(c + a)(c + b)≥ 3

4, which is trivial.

Remark.A stronger inequality is the following:

a

b + c(y + z) +

b

c + a(z + x) +

c

a + b(x + y) ≥

∑√(x + y)(x + z)− (x + y + z),

which may be obtained by applying the Cauchy-Schwarz Inequality, as follows

a

b + c(y + z) +

b

c + a(z + x) +

c

a + b(x + y) =

(a+b+c)(

y + z

b + c+

z + x

c + a+

x + y

a + b

)−2(x+y+z) ≥ 1

2(√

y + z+√

z + x+√

x + y)2−

2(x + y + z) =∑√

(x + y)(x + z)− (x + y + z).

A good exercise for readers is to show that∑√(x + y)(x + z) ≥ x + y + z +

√3(xy + yz + zx).


(b + c− a)2

(b + c)2 + a2+

(c + a− b)2

(c + a)2 + b2+

(a + b− c)2

(a + b)2 + c2≥ 3

5.

Japan, 1997

First solution:Let x =

b + c

a, y =

c + a

b, z =

a + b

c. The inequality can be written

∑ (x− 1)2

x2 + 1≥ 3

5.

Using the Cauchy-Schwarz Inequality, we find that∑ (x− 1)2

x2 + 1≥ (x + y + z − 3)2

x2 + y2 + z2 + 3

96 Solutions

and so it is enough to prove that

(x + y + z − 3)2

x2 + y2 + z2 + 3≥ 3

5⇔ (

∑x)2 − 15

∑x + 3

∑xy + 18 ≥ 0.

But from Schur’s Inequality, after some computations, we deduce that∑xy ≥ 2

∑x. Thus, we have(∑x)2

− 15∑

x + 3∑

xy + 18 ≥(∑

x)2

− 9∑

x + 18 ≥ 0,

the last one being clearly true since∑

x ≥ 6.

Second solution:Of course, we may take a + b + c = 2. The inequality becomes∑ 4(1− a)2

2 + 2(1− a)2≥ 3

5⇔

∑ 11 + (1− a)2

≤ 2710

.

But with the substitution 1 − a = x, 1 − b = y, 1 − c = z, the inequality reduces tothat from problem 47.

103. [ Vasile Cırtoaje, Gabriel Dospinescu ] Prove that if a1, a2, . . . , an ≥ 0 then

an1 + an

2 + · · ·+ ann − na1a2 . . . an ≥ (n− 1)

(a1 + a2 + · · ·+ an−1

n− 1− an

)n

where an is the least among the numbers a1, a2, . . . , an.

Solution:Let ai − an = xi ≥ 0 for i ∈ {1, 2, . . . , n− 1}. Now, let us look at

n∑i=1

ani − n ·

n∏i=1

ai − (n− 1)

n−1∑i=1

ai

n− 1− an

n

as a polynomial in a = an. It is in fact

an +n−1∑i=1

(a + xi)n − na

n−1∏i=1

(a + xi)− (n− 1)(

x1 + x2 + · · ·+ xn−1

n− 1

)n

.

We will prove that the coefficient of ak is nonnegative for all k ∈ {0, 1, . . . , n−1},because clearly the degree of this polynomial is at most n− 1. For k = 0, this followsfrom the convexity of the function f(x) = xn

n−1∑i=1

xni ≥ (n− 1)

n−1∑i=1

xi

n− 1

n

.


For k > 0, the coefficient of ak is(n

k

) n−1∑i=1

xn−ki − n

∑1≤i1<···<in−k≤n−1

xi1xi2 . . . xin−k.

Let us prove that this is nonnegative. From the AM-GM Inequality we have

n∑

1≤i1<i2<···<in−k≤n−1

xi1xi2 . . . xin−k≤

n

n− k

∑1≤i1<i2<···<in−k≤n−1

(xn−k

i1+ xn−k

i2+ · · ·+ xn−k

in−k

)=

k

n− 1·(

n

k

) n−1∑i=1

xn−ki

which is clearly smaller than(

n

k

) n−1∑i=1

xn−ki . This shows that each coefficient of the

polynomial is nonnegative and so this polynomial takes nonnegative values whenrestricted to nonnegative numbers.

104. [ Turkevici ] Prove that for all positive real numbers x, y, z, t,


Kvant

Solution:Clearly, it is enough to prove the inequality if xyzt = 1 and so the problem

becomesIf a, b, c, d have product 1, then a2 + b2 + c2 +d2 +2 ≥ ab+ bc+ cd+da+ac+ bd.

Let d the minimum among a, b, c, d and let m = 3√

abc. We will prove that

a2 + b2 + c2 + d2 + 2− (ab + bc + cd + da + ac + bd) ≥ d2 + 3m2 + 2− (3m2 + 3md),

which is in fact

a2 + b2 + c2 − ab− bc− ca ≥ d(a + b + c− 3 3

√abc)

.

Because d ≤ 3√

abc, proving this first inequality comes down to the inequality

a2 + b2 + c2 − ab− bc− ca ≥ 3√

abc(a + b + c− 3 3

√abc)

.

Take u =a

3√

abc, v =

b3√

abc, w =

c3√

abc. Using problem 74, we find that

u2 + v2 + w2 + 3 ≥ u + v + w + uv + vw + wu

which is exactly a2 + b2 + c2 − ab − bc − ca ≥ 3√

abc(a + b + c− 3 3

√abc). Thus, it

remains to prove that d2 + 2 ≥ 3md ⇔ d2 + 2 ≥ 3 3√

d2, which is clear.

98 Solutions

105. Prove that for any real numbers a1, a2, . . . , an the following inequality holds(n∑

i=1

ai

)2

≤n∑

i,j=1

ij

i + j − 1aiaj .

Solution:Observe that

n∑i,j=1

ij

i + j − 1aiaj =

n∑i,j=1

iai · jaj

∫ 1

0

ti+j−2dt =

=∫ 1

0

n∑i,j=1

iai · jaj · ti−1+j−1

dt =

=∫ 1

0

(n∑

i=1

iai · ti−1

)2

.

Now, using the Cauchy-Schwarz Inequality for integrals, we get∫ 1

0

(n∑

i=1

iai · ti−1

)2

dt ≥

(∫ 1

0

(n∑

i=1

iai · ti−1

)dt

)2

=

(n∑

i=1

ai

)2

,

which ends the proof.

106. Prove that if a1, a2, . . . , an, b1, . . . , bn are real numbers between 1001 and2002, inclusively, such that a2

1 + a22 + · · ·+ a2

n = b21 + b2

2 + · · ·+ b2n, then we have the

inequality

a31

b1+

a32

b2+ · · ·+ a3

n

bn≤ 17

10(a21 + a2

2 + · · ·+ a2n

).

TST Singapore

Solution:

The key ideas are thatai

bi∈[12, 2]

for any i and that for all x ∈[12, 2]

we have

the inequality x2 + 1 ≤ 52x. Consequently, we have

52· ai

bi≥ 1 +

a21

b2i

⇒ 52aibi ≥ a2

i + b2i

and also52

n∑i=1

aibi ≥n∑

i=1

(a2

i + b2i

)= 2

n∑i=1

a2i (1)


Now, the observation thata2

i

b2i

=

a3i

bi

ai · biand the inequality

52· ai

bi≥ 1 +

a2i

b2i

allow us to

write52a2

i ≥a3

i

bi+ aibi and adding up these inequalities yields

52

n∑i=1

a2i ≥

n∑i=1

(a3

i

bi+ aibi

)=

n∑i=1

a3i

bi+

n∑i=1

aibi (2)

Using (1) and (2) we find that

a31

b1+

a32

b2+ · · ·+ a3

n

bn≤ 17

10(a21 + a2

2 + · · ·+ a2n

),

which is the desired inequality.

107. [ Titu Andreescu, Gabriel Dospinescu ] Prove that if a, b, c are positive realnumbers which add up to 1, then

(a2 + b2)(b2 + c2)(c2 + a2) ≥ 8(a2b2 + b2c2 + c2a2)2.

Solution:Let x =

1a, y =

1b, z =

1c. We find the equivalent form if

1x

+1y

+1z

= 1 then

(x2 + y2)(y2 + z2)(z2 + x2) ≥ 8(x2 + y2 + z2)2.

We will prove the following inequality

(x2 + y2)(y2 + z2)(z2 + x2)(

1x

+1y

+1z

)2

≥ 8(x2 + y2 + z2)2

for any positive numbers x, y, z.Write x2 + y2 = 2c, y2 + z2 = 2a, z2 + x2 = 2b. Then the inequality becomes∑√

abc

b + c− a≥∑

a.

Recall Schur’s Inequality∑a4 + abc(a + b + c) ≥

∑a3(b + c) ⇔ abc(a + b + c) ≥

∑a3(b + c− a).

Now, using Holder’s Inequality, we find that

∑a3(b + c− a) =

∑ a3(1√

b + c− a

)2 ≥

(∑a)3

(∑ 1√b + c− a

)2 .

Combining the two inequalities, we find that∑√abc

b + c− a≥∑

a

and so the inequality is proved.

100 Solutions

108. [ Vasile Cırtoaje ] If a, b, c, d are positive real numbers such that abcd = 1,then

1(1 + a)2

+1

(1 + b)2+

1(1 + c)2

+1

(1 + d)2≥ 1.

Gazeta Matematica

Solution:If follows by summing the inequalities

1(1 + a)2

+1

(1 + b)2≥ 1

1 + ab,

1(1 + c)2

+1

(1 + d)2≥ 1

1 + cd.

The first from these inequalities follows from

1(1 + a)2

+1

(1 + b)2− 1

1 + ab=

ab(a2 + b2)− a2b2 − 2ab + 1(1 + a)2(1 + b)2(1 + c)2

=

=ab(a− b)2 + (ab− 1)2

(1 + a)2(1 + b)2(1 + ab)≥ 0.

Equality holds if a = b = c = d = 1.

109. [ Vasile Cırtoaje ] Let a, b, c be positive real numbers. Prove that

a2

b2 + c2+

b2

c2 + a2+

c2

a2 + b2≥ a

b + c+

b

c + a+

c

a + b.

Gazeta Matematica

Solution:We have the following identities

a2

b2 + c2− a

b + c=

ab(a− b) + ac(a− c)(b + c)(b2 + c2)

b2

c2 + a2− b

c + a=

bc(b− c) + ab(b− a)(c + a)(c2 + a2)

c2

a2 + b2− c

a + b=

ac(c− a) + bc(c− b)(b + a)(b2 + a2)

.

Thus, we have∑ a2

b2 + c2−∑ a

b + c=∑[

ab(a− b)(b + c)(b2 − c2)

− ab(a− b)(a + c)(a2 + c2)

]=

=(a2 + b2 + c2 + ab + bc + ca

)·∑ ab(a− b)2

(b + c)(c + a)(b2 + c2)(c2 + a2)≥ 0.


110. [ Gabriel Dospinescu ] Let a1, a2, . . . , an be real numbers and let S be anon-empty subset of {1, 2, . . . , n}. Prove that

(∑i∈S

ai

)2

≤∑

1≤i≤j≤n

(ai + . . . + aj)2.

TST 2004, Romania

First solution:Denote sk = a1 + · · ·+ ak for k = 1, n and also sn+1 = 0. Define now

bi =

{1 if i ∈ S

0 , otherwise

Using Abel’s summation we find that∑i∈S

ai = a1b1 + a2b2 + · · ·+ anbn =

= s1(b1 − b2) + s2(b2 − b3) + · · ·+ sn−1(bn−1 − bn) + snbn + sn+1(−b1)

Now, put b1 − b2 = x1, b2 − b3 = x2, . . . , bn−1 − bn = xn−1, bn = xn, xn+1 = −b1. Sowe have ∑

i∈S

ai =n+1∑i=1

xisi

and also xi ∈ {−1, 0, 1}. Clearly,n+1∑i=1

xi = 0. On the other hand, using Lagrange

identity we find that∑1≤i≤j≤n

(ai + · · ·+ aj)2 =

n∑i=1

s2i +

∑1≤i<j≤n

(sj − si)2 =

=∑

1≤i<j≤n+1

(sj − si)2 = (n + 1)n+1∑i=1

s2i −

(n+1∑i=1

si

)2

.

So we need to prove that

(n + 1)n+1∑i=1

s2i ≥

(n+1∑i=1

sixi

)2

+

(n+1∑i=1

si

)2

.

But it is clear that

(n+1∑i=1

sixi

)2

+

(n+1∑i=1

si

)2

=n+1∑i=1

s2i (1 + x2

i )

+ 2∑

1≤i<j≤n+1

sisj(xixj + 1)

102 Solutions

Now, using the fact that 2sisj ≤ si2 + s2

j , 1 + xixj ≥ 0, we can write

2∑

1≤i<jn+1

sisj(xixj + 1) ≤∑

1≤i<j≤n+1

(s2i + s2

j )(1 + xixj) =

n (s21 + · · ·+ s2

n+1) + s21x1(x2 + x3 + · · ·+ xn) + · · ·+ s2

nxn(x1 +

+ x2 + · · ·+ xn−1) = −s21x

21 − · · · − s2

nx2n + n(s2

1 + · · ·+ s2n).

So, (n+1∑i=1

sixi

)2

+

(n+1∑i=1

si

)2

≤n+1∑k=1

s2k(x2

k + 1) +n+1∑k=1

s2k(n− x2

k) =

= (n + 1)n+1∑k+1

s2k and we are done.

Second solution (by Andrei Negut):First, let us prove a lemma

LemmaFor any a1, a2, . . . , a2k+1 ∈ R we have the inequality(

k∑i=0

a2i+1

)2

≤∑

1≤i≤j≤2k+1

(ai + · · ·+ aj)2.

Proof of the lemmaLet us take sk = a1 + · · ·+ ak. We have

k∑i=0

a2i+1 = s1 + s3 − s2 + · · ·+ s2k+1 − s2k

and so the left hand side in the lemma is2k+1∑i=1

s2i + 2

∑0≤i<j≤k

s2i+1s2j+1 + 2∑

1≤i<j≤k

s2is2j − 2∑

0≤i≤k1≤j≤k

s2i+1s2j

and the right hand side is just

(2k + 1)2k+1∑i=1

s2i − 2

∑1≤i<j≤2k+1

sisj .

Thus, we need to prove that

2k2k+1∑i=1

s2i ≥ 4

∑0≤i<j≤k

s2i+1s2j+1 + 4∑

1≤i<j≤k

s2is2j

and it comes by adding up the inequalities

2s2j+1s2i+1 ≤ s22i+1 + s2

2j+1, 2s2is2j ≤ s22i + s2

2j .


Now, let us turn back to the solution of the problem. Let us call a successionof ai’s a sequence and call a sequence that is missing from S a gap. We group thesuccessive sequences from S and thus S will look like this

S = {ai1 , ai1+1, . . . , ai1+k1 , ai2 , ai2+1, . . . , ai2+k2 , . . . , air, . . . , air+kr

}

where ij + kj < ij+1 − 1. Thus, we write S as a sequence, followed by a gap, followedby a sequence, then a gap and so on. Now, take s1 = ai1 + · · ·+ai1+k1 , s2 = ai1+k1+1+· · ·+ ai2−1, . . . , s2r−1 = air

+ · · ·+ air+kr

Then,(∑i∈S

ai

)2

= (s1 + s3 + · · ·+ s2r−1)2 ≤

∑1≤i≤j≤2r−1

(si + · · ·+ sj)2 ≤

∑1≤i≤j≤n

(ai + · · ·+ aj)2

the last inequality being clearly true, because the terms in the left hand side areamong those from the right hand side.

Third solution:We will prove the inequality using induction. For n = 2 and n = 3 it’s easy.

Suppose the inequality is true for all k < n and let us prove it for n. If 1 /∈ S, thenwe just apply the inductive step for the numbers a2, . . . , an, because the right handside doesn’t decrease. Now, suppose 1 ∈ S. If 2 ∈ S, then we apply the inductive stepwith the numbers a1 + a2, a3, . . . , an. Thus, we may assume that 2 /∈ S. It is easy to

see that (a + b + c)2 + c2 ≥ (a + b)2

2≥ 2ab and thus we have (a1 + a2 + · · ·+ ak)2 +

(a2 +a3 + · · ·+ak−1)2 ≥ 2a1ak (*). Also, the inductive step for a3, . . . , an shows that ∑i∈S\{1}

2

≤∑

3≤i≤j≤n

(ai + · · ·+ aj)2.

So, it suffices to show that

a21 + 2a1

∑i∈S\{1}

ai ≤n∑

i=1

(a1 + · · ·+ ai)2 +

n∑i=2

(a2 + · · ·+ ai)2

But this is clear from the fact that a21 appears in the right hand side and by summing

up the inequalities from (*).

111. [ Dung Tran Nam ] Let x1, x2 . . . , x2004 be real numbers in the interval [−1, 1]such that x3

1 +x32 + . . .+x3

2004 = 0. Find the maximal value of the x1 +x2 + · · ·+x2004.

Solution:Let us take ai = x3

i and the function f : [−1, 1] → R, f(x) = x13 . We will prove

first the following properties of f :

104 Solutions

1. f(x + y + 1) + f(−1) ≥ f(x) + f(y) if −1 < x, y < 0.2. f is convex on [−1, 0] and concave on [0, 1].3. If x > 0 and y < 0 and x + y < 0 then f(x) + f(y) ≤ f(−1) + f(x + y + 1) and ifx + y > 0 then f(x) + f(y) ≤ f(x + y). The proofs of these results are easy. Indeed,for the first one we make the substitution x = −a3, y = −b3 and it comes down to1 > a3+b3+(1−a−b)3 ⇔ 1 > (a+b)(a2−ab+b2)+1−3(a+b)+3(a+b)2−(a+b)3 ⇔⇔ 3(a + b)(1 − a)(1 − b) > 0, which follows. The second statement is clear and thethird one can be easily deduced in the same manner as 1.

From these arguments we deduce that if (t1, t2, . . . , t2004) = t is the point wherethe maximum value of the function g : A = {x ∈ [−1, 1]2004|x1 + · · ·+ xn = 0} →

→ R, g(x1, . . . , x2004) =2004∑k=1

f(xk) (this maximum exists because this function is

defined on a compact) is attained then we have that all positive components of t areequal to each other and all negative ones are −1. So, suppose we have k componentsequal to−1 and 2004−k components equal to a number a. Because t1+t2+· · ·+t2004 =

0 we find that a =k

2004− kand the value of g in this point is (2004−k) 3

√k

2004− k−k.

Thus we have to find the maximum value of (2004− k) 3

√k

2004− k− k, when k is in

the set {0, 1, . . . , 2004}. A short analysis with derivatives shows that the maximum isattained when k = 223 and so the maximum value is 3

√223 · 3

√17812 − 223.

112. [ Gabriel Dospinescu, Calin Popa ] Prove that if n ≥ 2 and a1, a2, . . . , an

are real numbers with product 1, then

a21 + a2

2 + · · ·+ a2n − n ≥ 2n

n− 1· n√

n− 1(a1 + a2 + · · ·+ an − n).

Solution:We will prove the inequality by induction. For n = 2 it is trivial. Now, suppose the

inequality is true for n−1 numbers and let us prove it for n. First, it is easy to see thatit is enough to prove it for a1, . . . , an > 0 (otherwise we replace a1, a2, . . . , an with|a1|, |a2|, . . . , |an|, which have product 1. Yet, the right hand side increases). Now,let an the maximum number among a1, a2, . . . , an and let G the geometric mean ofa1, a2, . . . , an−1. First, we will prove that

a21 + a2

2 + · · ·+ a2n − n− 2n

n− 1· n√

n− 1(a1 + a2 + · · ·+ an − n) ≥

≥ a2n + (n− 1)G2 − n− 2n

n− 1· n√

n− 1(an + (n− 1)G− n)

which is equivalent to

a21 + a2

2 + · · ·+ a2n−1 − (n− 1) n−1

√a21a

22 . . . a2

n−1 ≥


≥ 2n

n− 1· n√

n− 1(a1 + a2 + · · ·+ an−1 − (n− 1) n−1

√a1a2 . . . an−1

).

Because, n−1√

a1a2 . . . an−1 ≤ 1 and a1+a2+· · ·+an−1−(n−1) n−1√

a1a2 . . . an−1 ≥0, it is enough to prove the inequality

a21 + · · ·+ a2

n−1 − (n− 1)G2 ≥ 2n

n− 1· n√

n− 1 ·G · (a1 + · · ·+ an−1 − (n− 1)G).

Now, we apply the inductive hypothesis for the numbersa1

G, . . . ,

an−1

Gwhich have

product 1 and we infer that

a21 + · · ·+ a2

n−1

G2− n + 1 ≥ 2(n− 1)

n− 2· n−1

√n− 2

(a1 + · · ·+ an−1

G− n + 1

)and so it suffices to prove that

2(n− 1)n− 2

· n−1√

n− 2(a1+· · ·+an−1−(n−1)G) ≥ 2n

n− 1· n√

n− 1(a1+· · ·+an−1−(n−1)G),

which is the same as 1 +1

n(n− 2)≥

n√

n− 1n−1√

n− 2. This becomes

(1 +

1n(n− 2)

)n(n−1)

≥ (n− 1)n−1

(n− 2)n

and it follows for n > 4 from(1 +

1n(n− 2)

)n(n−1)

> 2

and

(n− 1)n−1

(n− 2)n=

1n− 2

·(

1 +1

n− 2

)·(

1 +1

n− 2

)n−2

<e

n− 2

(1 +

1n− 2

)< 2.

For n = 3 and n = 4 it is easy to check.Thus, we have proved that

a21 + a2

2 + · · ·+ a2n − n− 2n

n− 1· n√

n− 1(a1 + a2 + · · ·+ an − n) ≥

≥ a2n + (n− 1)G2 − n− 2n

n− 1· n√

n− 1(an + (n− 1)G− n)

and it is enough to prove that

x2(n−1) +n− 1x2

− n ≥ 2n

n− 1· n√

n− 1(

xn−1 +n− 1

x− n

)for all x ≥ 1 (we took x =

1G

). Let us consider the function

f(x) = x2(n−1) +n− 1x2

− n− 2n

n− 1· n√

n− 1(

xn−1 +n− 1

x− n

).

We have

f ′(x) = 2 · xn − 1x2

·[(n− 1)(xn + 1)

x− n n

√n− 1

]≥ 0

106 Solutions

because

xn−1 +1x

= xn−1 +1

(n− 1)x+ · · ·+ 1

(n− 1)x≥ n n

√1

(n− 1)n−1.

Thus, f is increasing and so f(x) ≥ f(1) = 0. This proves the inequality.

113. [ Vasile Cırtoaje ] If a, b, c are positive real numbers, then√2a

a + b+

√2b

b + c+

√2c

c + a≤ 3.

Gazeta Matematica

First solution:

With the notations x =

√b

a, y =

√c

b, z =

√a

cthe problem reduces to proving

that xyz = 1 implies √2

1 + x2+√

21 + y2

+

√2

1 + z2≤ 3.

We presume that x ≤ y ≤ z, which implies xy ≤ 1 and z ≥ 1. We have(√2

1 + x2+√

21 + y2

)2

≤ 2(

21 + x2

+2

1 + y2

)= 4

[1 +

1− x2y2

(1 + x2)(1 + y2)

]≤

4[1 +

1− x2y2

(1 + xy)2

]=

81 + xy

=8z

z + 1.

so √2

1 + x2+√

21 + y2

≤ 2

√2z

z + 1and we need to prove that

2

√2z

z + 1+

√2

1 + z2≤ 3.

Because √2

1 + z2≤ 2

1 + zwe only need to prove that

2

√2z

z + 1+

21 + z

≤ 3.

This inequality is equivalent to

1 + 3z − 2√

2z(1 + z) ≥ 0, (√

2z −√

z + 1)2 ≥ 0,

and we are done.


Second solution:Clearly, the problem asks to prove that if xyz = 1 then∑√

2x + 1

≤ 3.

We have two cases. The first and easy one is when xy + yz + zx ≥ x + y + z. In thiscase we can apply the Cauchy-Schwarz Inequality to get∑√

2x + 1

≤√

3∑ 2

x + 1.

But∑ 1x + 1

≤ 32

⇔∑

2(xy + x + y + 1) ≤ 3(2 + x + y + z + xy + yz + zx) ⇔

⇔ x + y + z ≤ xy + yz + zx

and so in this case the inequality is proved.The second case is when xy + yz + zx < x + y + z. Thus,

(x− 1)(y − 1)(z − 1) = x + y + z − xy − yz − zx > 0

and so exactly two of the numbers x, y, z are smaller than 1, let them be x and y. So,we must prove that if x and y are smaller than 1, then√

2x + 1

+√

2y + 1

+√

2xy

xy + 1≤ 3.

Using the Cauchy-Schwarz Inequality, we get√2

x + 1+√

2y + 1

+√

2xy

xy + 1≤ 2√

1x + 1

+1

y + 1+√

2xy

xy + 1

and so it is enough to prove that this last quantity is at most 3. But this comes downto

2 ·

1x + 1

+1

y + 1− 1

1 +√

1x + 1

+1

y + 1

≤1− 2xy

xy + 1

1 +√

2xy

xy + 1

.

Because we have1

x + 1+

1y + 1

≥ 1, the left hand side is at most

1x + 1

+1

y + 1− 1 =

1− xy

(x + 1)(y + 1)

and so we are left with the inequality

1− xy

(x + 1)(y + 1)≤ 1− xy

(xy + 1)(

1 +√

2xy

xy + 1

) ⇔ xy + 1 + (xy + 1)√

2xy

xy + 1≤

≤ xy + 1 + x + y ⇔ x + y ≥√

2xy(xy + 1)

which follows from√

2xy(xy + 1) ≤ 2√

xy ≤ x + y. The problem is solved.

108 Solutions

114. Prove the following inequality for positive real numbers x, y, z

(xy + yz + zx)(

1(x + y)2

+1

(y + z)2+

1(z + x)2

)≥ 9

4

Iran, 1996

First solution (by Iurie Boreico):With the substitution x+y = c, y +z = a, z +x = b, the inequality becomes after

some easy computations ∑(2ab− 1

c2

)(a− b)2 ≥ 0.

Let a ≥ b ≥ c. If 2c2 ≥ ab, each term in the above expression is positive andwe are done. So, let 2c2 < ab. First, we prove that 2b2 ≥ ac, 2a2 ≥ bc. Suppose that2b2 < ac. Then (b + c)2 ≤ 2(b2 + c2) < a(b + c) and so b + c < a, false. Clearly, wecan write the inequality like that(

2ac− 1

b2

)(a− c)2 +

(2bc− 1

a2

)(b− c)2 ≥

(1c2− 2

ab

)(a− b)2.

We can immediately see that the inequality (a − c)2 ≥ (a − b)2 + (b − c)2 holdsand thus if suffices to prove that(

2ac

+2bc− 1

a2− 1

b2

)(b− c)2 ≥

(1b2

+1c2− 2

ab− 2

ac

)(a− b)2.

But it is clear that1b2

+1c2− 2

ab− 2

ac<

(1b− 1

c

)2

and so the right hand side

is at most(a− b)2(b− c)2

b2c2. Also, it is easy to see that

2ac

+2bc− 1

a2− 1

b2≥ 1

ac+

1bc

>(a− b)2

b2c2,

which shows that the left hand side is at least(a− b)2(b− c)2

b2c2and this ends the

solution.

Second solution:Since the inequality is homogenous, we may assume that xy + yz + zx = 3. Also,

we make the substitution x + y + z = 3a. From (x + y + z)2 ≥ 3(xy + yz + zx), weget a ≥ 1. Now we write the inequality as follows

1(3a− z)2

+1

(3a− y)2+

1(3a− x)2

≥ 34,

4[(xy + 3az)2 + (yz + 3ax)2 + (zx + 3ay)2] ≥ 3(9a− xyz)2,

4(27a4 − 18a2 + 3 + 4axyz) ≥ (9a− xyz)2,

3(12a2 − 1)(3a2 − 4) + xyz(34a− xyz) ≥ 0, (1)


12(3a2 − 1)2 + 208a2 ≥ (17a− xyz)2. (2)

We have two cases.i) Case 3a2 − 4 ≥ 0. Since

34a− xyz =19[34(x + y + z)(xy + yz + zx)− 9xyz] > 0,

the inequality (1) is true.ii) Case 3a2 − 4 < 0. From Schur’s Inequality

(x + y + z)3 − 4(x + y + z)(xy + yz + zx) + 9xyz ≥ 0,

it follows that 3a3 − 4a + xyz ≥ 0. Thus,

12(3a2 − 1)2 + 208a2 − (17a− xyz)2 ≥ 12(3a2 − 1)2 + 208a2 − a2(3a2 + 13)2 =

= 3(4− 11a2 + 10a4 − 3a6) = 3(1− a2)2(4− 3a2)2 ≥ 0.

115. Prove that for any x, y in the interval [0, 1],√1 + x2 +

√1 + y2 +

√(1− x)2 + (1− y)2 ≥ (1 +

√5)(1− xy).

Solution (by Faruk F. Abi-Khuzam and Roy Barbara - ”A sharp in-

equality and the iradius conjecture”):Let the function F : [0, 1]2 → R, F (x, y) =

√1 + x2 +

√1 + y2 +√

(1− x)2 + (1− y)2 − (1 +√

5)(1 − xy). It is clear that F is symmetric in x andy and also the convexity of the function x →

√1 + x2 shows that F (x, 0) ≥ 0 for all

x. Now, suppose we fix y and consider F as a function in x. It’s derivatives are

f ′(x) = (1 +√

5)y +x√

x2 + 1− 1− x√

(1− x)2 + (1− y)2

and

f ′′(x) =1√

(1 + x2)3+

(1− y)2√((1− x)2 + (1− y)2)3

.

Thus, f is convex and its derivative is increasing. Now, let

r =

√1− 6

√3

(1 +√

5)2, c = 1 +

√5.

The first case we will discuss is y ≥ 14. It is easy to see that in this case we have cy >

1√y2 − 2y + 2

(the derivative of the function y → y2c2(y2 − 2y + 2)− 1 is positive)

and so f ′(0) > 0. Because f ′ is increasing, we have f ′(x) > 0 and so f is increasingwith f(0) = F (0, y) = F (y, 0) ≥ 0. Thus, in this case the inequality is proved.

110 Solutions

Due to case 1 and to symmetry, it remains to show that the inequality holds in

the cases x ∈[0,

14

], y ∈ [0, r] and x ∈

[r,

14

], y ∈

[r,

14

].

In the first case we deduce immediately that

f ′(x) ≤ r(1 +√

5) +x√

1 + x2− 1− x√

1 + (x− 1)2

Thus, we have f ′(

14

)< 0, which shows that f is decreasing on

[0,

14

]. Because from

the first case discussed we have f

(14

)≥ 0, we will have F (x, y) = f(x) ≥ 0 for all

points (x, y) with x ∈[0,

14

], y ∈ [0, r].

Now, let us discuss the most important case, when x ∈[r,

14

], y ∈

[r,

14

]. Let

the points O(0, 0), A(1, 0), B(1, 1), C(0, 1),M(1, y), N(x, 1). The triangle OMN hasperimeter √

1 + x2 +√

1 + y2 +√

(1− x)2 + (1− y)2 and area1− xy

2.

But it is trivial to show that in any triangle with perimeter P and area S we have the

inequality S ≤ P 2

12√

3. Thus, we find that

√1 + x2+

√1 + y2+

√(1− x)2 + (1− y)2 ≥√

6√

3√

1− xy ≥ (1+√

5)(1−xy) due to the fact that xy ≥ r2. The proof is complete.

116. [ Suranyi ] Prove that for any positive real numbers a1, a2, . . . , an the fol-lowing inequality holds

(n−1)(an1 +an

2 +· · ·+ann)+na1a2 . . . an ≥ (a1+a2+· · ·+an)(an−1

1 +an−12 +· · ·+an−1

n ).

Miklos Schweitzer Competition

Solution:Again, we will use induction to prove this inequality, but the proof of the inductive

step will be again highly non-trivial. Indeed, suppose the inequality is true for n

numbers and let us prove it for n + 1 numbers.Due to the symmetry and homogeneity of the inequality, it is enough to prove it

under the conditions a1 ≥ a2 ≥ · · · ≥ an+1 and a1 + a2 + · · · + an = 1. We have toprove that

nn∑

i=1

an+1i + nan+1

n+1 + nan+1 ·n∏

i=1

ai + an+1 ·n∏

i=1

ai − (1 + an+1)

(n∑

i=1

ani + an

n+1

)≥ 0.

But from the inductive hypothesis we have

(n− 1)(an1 + an

2 + · · ·+ ann) + na1a2 . . . an ≥ an−1

1 + an−12 + · · ·+ an−1

n


and so

nan+1

n∏i=1

ai ≥ an+1

n∑i=1

an−1i − (n− 1)an+1

n∑i=1

ani .

Using this last inequality, it remains to prove that(n

n∑i=1

an+1i −

n∑i=1

ani

)− an+1

(n

n∑i=1

ani −

n∑i=1

an−1i

)+

+an+1

(n∏

i=1

ai + (n− 1)ann+1 − an−1

n+1

)≥ 0.

Now, we will break this inequality into

an+1

(n∏

i=1

ai + (n− 1)ann+1 − an−1

n+1

)≥ 0

and (n

n∑i=1

an+1i −

n∑i=1

ani

)− an+1

(n

n∑i=1

ani −

n∑i=1

an−1i

)≥ 0.

Let us justify these two inequalities. The first one is pretty obviousn∏

i=1

ai + (n− 1)ann+1 − an−1

n+1 =n∏

i=1

(ai − an+1 + an+1) + (n− 1)ann+1 − an−1

n+1 ≥

≥ ann+1 + an−1

n+1 ·n∑

i=1

(ai − an+1) + (n− 1)ann+1 − an−1

n+1 = 0.

Now, let us prove the second inequality. It can be written as

nn∑

i=1

an+1i −

n∑i=1

ani ≥ an+1

(n

n∑i=1

ani −

n∑i=1

an−1i

).

Because nn∑

i=1

ani −

n∑i=1

an−1i ≥ 0 (using Chebyshev’s Inequality) and an+1 ≤

1n

, it

is enough to prove that

nn∑

i=1

an+1i −

n∑i=1

ani ≥

1n

(n

n∑i=1

ani −

n∑i=1

an−1i

).

but this one follows, because nan+1i +

1n

an−1i ≥ an

i for all i. Thus, the inductive stepis proved.

117. Prove that for any x1, x2, . . . , xn > 0 with product 1,∑1≤i<j≤n

(xi − xj)2 ≥n∑

i=1

x2i − n.

A generalization of Turkevici’s inequality

112 Solutions

Solution:Of course, the inequality can be written in the following form

n ≥ −(n− 1)

(n∑

k=1

x2k

)+

(n∑

k=1

xk

)2

.

We will prove this inequality by induction. The case n = 2 is trivial. Suppose theinequality is true for n− 1 and let us prove it for n. Let

f(x1, x2, . . . , xn) = −(n− 1)

(n∑

k=1

x2k

)+

(n∑

k=1

xk

)2

and let G = n−1√

x2x3 . . . xn, where we have already chosen x1 = min{x1, x2, . . . , xn}.It is easy to see that the inequality f(x1, x2, . . . , xn) ≤ f(x1, G,G, . . . , G) is equivalentto

(n− 1)n∑

k=2

x2k −

(n∑

k=2

xk

)2

≥ 2x1

(n∑

k=2

xk − (n− 1)G

).

Clearly, we have x1 ≤ G andn∑

k=2

xk ≥ (n− 1)G, so it suffices to prove that

(n− 1)n∑

k=2

x2k −

(n∑

k=2

xk

)2

≥ 2G

(n∑

k=2

xk − (n− 1)G

).

We will prove that this inequality holds for all x2, . . . , xn > 0. Because the inequalityis homogeneous, it is enough to prove it when G = 1. In this case, from the inductionstep we already have

(n− 2)n∑

k=2

x2k + n− 1 ≥

(n∑

k=2

xk

)2

and so it suffices to prove thatn∑

k=2

x2k + n− 1 ≥

n∑k=2

2xk ⇔n∑

k=2

(xk − 1)2 ≥ 0,

clearly true.Thus, we have proved that f(x1, x2, . . . , xn) ≤ f(x1, G,G, . . . , G). Now, to com-

plete the inductive step, we will prove that f(x1, G,G, . . . , G) ≤ 0. Because clearly

x1 =1

Gn−1, the last assertion reduces to proving that

(n− 1)(

(n− 1)G2 +1

G2(n−1)

)+ n ≥

((n− 1)G +

1Gn−1

)2

which comes down ton− 2G2n−2

+ n ≥ 2n− 2Gn−2

and this one is an immediate consequence of the AM-GM Inequality.


118. [ Gabriel Dospinescu ] Find the minimum value of the expressionn∑

i=1

√a1a2 . . . an

1− (n− 1)ai

where a1, a2, . . . , an <1

n− 1add up to 1 and n > 2 is an integer.

Solution:We will prove that the minimal value is

1√nn−3

. Indeed, using Suranyi’s In-

equality, we find that

(n− 1)n∑

i=1

ani + na1a2 . . . an ≥

n∑i=1

an−1i ⇒ na1a2 . . . an ≥

n∑i=1

an−1i (1− (n− 1)ai).

Now, let us observe that

n∑k=1

an−1k (1− (n− 1)ak) =

n∑k=1

(3

√an−1

k

)3

(√1

1− (n− 1)ak

)2

and so, by an immediate application of Holder Inequality, we have

n∑i=1

an−1i (1− (n− 1)ai) ≥

(n∑

k=1

an−1

3k

)3

(n∑

k=1

√1

1− (n− 1)ak

)2 .

But for n > 3, we can apply Jensen’s Inequality to deduce that

n∑k=1

an−1

3k

n≥

n∑

k=1

ak

n

n−1

3

=1

nn−1

3

.

Thus, combining all these inequalities, we have proved the inequality for n > 3.For n = 3 it reduces to proving that∑√

abc

b + c− a≥∑

a

which was proved in the solution of the problem 107.

119. [ Vasile Cırtoaje ] Let a1, a2, . . . , an < 1 be nonnegative real numbers suchthat

a =

√a21 + a2

2 + . . . + a2n

n≥√

33

.

114 Solutions

Prove thata1

1− a21

+a2

1− a22

+ . . . +an

1− a2n

≥ na

1− a2.

Solution:We proceed by induction. Clearly, the inequality is trivial for n= 1. Now we

suppose that the inequality is valid for n = k − 1, k ≥ 2 and will prove that

a1

1− a21

+a2

1− a22

+ . . . +ak

1− a2k

≥ ka

1− a2,

for

a =

√a21 + a2

2 + . . . + a2k

k≥√

33

.

We assume, without loss of generality, that a1 ≥ a2 ≥ . . . ≥ ak, therefore a ≥ ak.

Using the notation

x =

√a21 + a2 + · · ·+ a2

k−1

k − 1=

√ka2 − a2

k

k − 1,

from a ≥ ak it follows that x ≥ a ≥√

33

. Thus, by the inductive hypothesis, we have

a1

1− a21

+a2

1− a22

+ . . . +ak−1

1− a2k−1

≥ (k − 1)x1− x2

,

and it remains to show thatak

1− a2k

+(k − 1)x1− x2

≥ ka

1− a2.

From

x2 − a2 =ka2 − a2

k

k − 1− a2 =

a2 − a2k

k − 1,

we obtain

(k − 1)(x− a) =(a− ak)(a + ak)

x + aand

ak

1− a2k

+(k − 1)x1− x2

− ka

1− a2=(

ak

1− a2k

− a

1− a2

)+ (k − 1)

(x

1− x2− a

1− a2

)=

−(a− ak)(1 + aak)(1− a2

k)(1− a2)+

(k − 1)(x− a)(1 + ax)(1− x2)(1− a2)

=a− ak

1− a2

[−(1 + aak)

1− a2k

+(a + ak)(1 + ax)(1− x2)(x + a)

]=

=(a− ak)(x− ak)[−1 + x2 + a2

k + xak + a(x + ak) + a2 + axak(x + ak) + a2xak](1− a2)(1− a2

k)(1− x2)(x + a).

Since x2 − a2k =

ka2 − a2k

k − 1− a2

k =k(a− ak)(a + ak)

k − 1, it follows that

(a− ak)(x− ak) =k(a− ak)2(a + ak)

(k − 1)(x + ak)≥ 0,


and hence we have to show that

x2 + a2k + xak + a(x + ak) + a2 + axak(x + ak) + a2xak ≥ 1.

In order to show this inequality, we notice that

x2 + a2k =

ka2 + (k − 2)a2k

k − 1≥ ka2

k − 1and

x + ak ≥√

x2 + a2k ≥ a

√k

k − 1.

Consequently, we have

x2 + a2k +xak + a(x+ ak) + a2 + axak(x+ ak) + a2xak ≥ (x2 + a2

k) + a(x+ ak) + a2 ≥

≥

(k

k − 1+

√k

k − 1+ 1

)a2 > 3a2 = 1,

and the proof is complete. Equality holds when a1 = a2 = . . . = an.

Remarks.1. From the final solution, we can easily see that the inequality is valid for the

larger condition

a ≥ 1√1 +

√n

n−1 + nn−1

.

This is the largest range for a, because for a1 = a2 = · · · = an−1 = x and an = 0

(therefore a = x

√n− 1

n), from the given inequality we get a ≥ 1√

1 +√

nn−1 + n

n−1

.

2. The special case n = 3 and a =√

33

is a problem from Crux 2003.

120. [ Vasile Cırtoaje, Mircea Lascu ] Let a, b, c, x, y, z be positive real numberssuch that

(a + b + c)(x + y + z) = (a2 + b2 + c2)(x2 + y2 + z2) = 4.

Prove thatabcxyz <

136

.

Solution:Using the AM-GM Inequality, we have

4(ab+bc+ca)(xy+yz+zx) =[(a + b + c)2 − (a2 + b2 + c2)

] [(x + y + z)2 − (x2 + y2 + z2)

]=

= 20− (a + b + c)2(x2 + y2 + z2)− (a2 + b2 + c2)(x + y + z)2 ≤

≤ 20− 2√

(a + b + c)2(x2 + y2 + z2)(a2 + b2 + c2)(x + y + z)2 = 4,

116 Solutions

therefore

(ab + bc + ca)(xy + yz + za) ≤ 1. (1)

By multiplying the well-known inequalities

(ab + bc + ca)2 ≥ 3abc(a + b + c), (xy + yz + zx)2 ≥ 3xyz(x + y + z),

it follows that

(ab + bc + ca)2(xy + yz + zx)2 ≥ 9abcxyz(a + b + c)(x + y + z),

or

(ab + bc + ca)(xy + yz + zx) ≥ 36abcxyz. (2)

From (1) and (2), we conclude that

1 ≤ (ab + bc + ca)(xy + yz + zx) ≥ 36abcxyz,

therefore 1 ≤ 36abcxyz.To have 1 = 36abcxyz, the equalities (ab + bc + ca)2 = 3abc(a + b + c) and

(xy +yz +zx)2 = 3xyz(x+y +z) are necessary. But these conditions imply a = b = c

and x = y = z, which contradict the relations (a + b + c)(x + y + z) = (a2 + b2 +c2)(x2 + y2 + z2) = 4. Thus, it follows that 1 > 36abcxyz.

121. [ Gabriel Dospinescu ] For a given n > 2, find the minimal value of theconstant kn, such that if x1, x2, . . . , xn > 0 have product 1, then

1√1 + knx1

+1√

1 + knx2

+ · · ·+ 1√1 + knxn

≤ n− 1.

Mathlinks Contest

Solution:We will prove that kn =

2n− 1(n− 1)2

. Taking x1 = x2 = · · · = xn = 1, we find that

kn ≥2n− 1

(n− 1)2. So, it remains to show that

n∑k=1

1√1 +

2n− 1(n− 1)2

xk

≤ n− 1

if x1 · x2 · · · · · xn = 1. Suppose this inequality doesn’t hold for a certain system of n

numbers with product 1, x1, x2, . . . , xn > 0. Thus, we can find a number M > n− 1and some numbers ai > 0 which add up to 1, such that

1√1 +

2n− 1(n− 1)2

xk

= Mak.


Thus, ak <1

n− 1and we have

1 =n∏

k=1

[(n− 1)2

2n− 1

(1

M2a2k

− 1)]

⇒(

2n− 1(n− 1)2

)n

<n∏

k=1

(1

(n− 1)2a2k

− 1)

.

Now, denote 1 − (n − 1)ak = bk > 0 and observe thatn∑

k=1

bk = 1. Also, the above

inequality becomes

(n− 1)2nn∏

k=1

bk ·n∏

k=1

(2− bk) > (2n− 1)n

(n∏

k=1

(1− bk)

)2

.

Because from the AM-GM Inequality we haven∏

k=1

(2− bk) ≤(

2n− 1n

)n

,

our assumption leads ton∏

k=1

(1− bk)2 <(n− 1)2n

nnb1b2 . . . bn.

So, it is enough to prove that for any positive numbers a1, a2, . . . , an the inequalityn∏

k=1

(a1 +a2 + · · ·+ak−1 +ak+1 + · · ·+an)2 ≥ (n− 1)2n

nna1a2 . . . an(a1 +a2 + · · ·+an)n

holds.This strong inequality will be proved by induction. For n = 3, it follows from the

fact that (∏(a + b)

)2

abc≥

(89·(∑

a)·(∑

ab))2

abc≥ 64

27

(∑a)3

.

Suppose the inequality is true for all systems of n numbers. Let a1, a2, . . . , an+1

be positive real numbers. Because the inequality is symmetric and homogeneous, wemay assume that a1 ≤ a2 ≤ · · · ≤ an+1 and also that a1 +a2 + · · ·+an = 1. Applyingthe inductive hypothesis we get the inequality

n∏i=1

(1− ai)2 ≥(n− 1)2n

nna1a2 . . . an.

To prove the inductive step, we must prove thatn∏

i=1

(an+1 + 1− ai)2 ≥n2n+2

(n + 1)n+1a1a2 . . . anan+1(1 + an+1)n+1.

Thus, it is enough to prove the stronger inequalityn∏

i=1

(1 +

an+1

1− ai

)2

≥ n3n+2

(n− 1)2n · (n + 1)n+1an+1(1 + an+1)n+1.

118 Solutions

Now, using Huygens Inequality and the AM-GM Inequality, we find that

n∏i=1

(1 +

an+1

1− ai

)2

≥

1 +an+1

n

√√√√ n∏i=1

(1− ai)

2n

≥(

1 +nan+1

n− 1

)2n

and so we are left with the inequality(1 +

nan+1

n− 1

)2n

≥ n3n+2

(n− 1)2n · (n + 1)n+1an+1(1 + an+1)n+1

if an+1 ≥ max{a1, a2, . . . , an} ≥1n

. So, we can putn(1 + an+1)

n + 1= 1 + x, where x is

nonnegative. So, the inequality becomes(1 +

x

n(x + 1)

)2n

≥ 1 + (n + 1)x(x + 1)n−1

.

Using Bernoulli Inequality, we find immediately that(1 +

x

n(x + 1)

)2n

≥ 3x + 1x + 1

.

Also, (1 + x)n−1 ≥ 1 + (n− 1)x and so it is enough to prove that

3x + 1x + 1

≥ 1 + (n + 1)x1 + (n− 1)x

which is trivial.So, we have reached a contradiction assuming the inequality doesn’t hold for a

certain system of n numbers with product 1, which shows that in fact the inequality

is true for kn =2n− 1

(n− 1)2and that this is the value asked by the problem.

Remark.For n = 3, we find an inequality stronger than a problem given in China Math-

ematical Olympiad in 2003. Also, the case n = 3 represents a problem proposed byVasile Cartoaje in Gazeta Matematica, Seria A.

122. [ Vasile Cırtoaje, Gabriel Dospinescu ] For a given n > 2, find the maximalvalue of the constant kn such that for any x1, x2, . . . , xn > 0 for which x2

1 +x22 + · · ·+

x2n = 1 we have the inequality

(1− x1)(1− x2) . . . (1− xn) ≥ knx1x2 . . . xn.


Solution:We will prove that this constant is (

√n− 1)n. Indeed, let ai = x2

i . We must findthe minimal value of the expression

n∏i=1

(1−√

ai)

n∏i=1

√ai

when a1+a2+· · ·+an = 1. Let us observe that proving that this minimum is (√

n−1)n

reduces to proving thatn∏

i=1

(1− ai) ≥ (√

n− 1)n ·n∏

i=1

√ai ·

n∏i=1

(1 +√

ai).

But from the result proved in the solution of the problem 121, we find thatn∏

i=1

(1− ai)2 ≥(

(n− 1)2

n

)n n∏i=1

ai.

So, it is enough to prove thatn∏

i=1

(1 +√

ai) ≤(

1 +1√n

)n

.

But this is an easy task, because from the AM-GM Inequality we get

n∏i=1

(1 +√

ai) ≤

1 +

n∑i=1

√ai

n

n

≤(

1 +1√n

)n

,

the last one being a simple consequence of the Cauchy-Schwarz Inequality.Remark.The case n = 4 was proposed by Vasile Cartoaje in Gazeta Matematica Annual

Contest, 2001.

Glossary

(1) Abel’s Summation FormulaIf a1, a2, ..., an, b1, b2, ..., bn are real or complex numbers, and

Si = a1 + a2 + ... + ai,i = 1, 2, ..., n,

thenn∑

i=1

aibi =n−1∑i=1

Si(bi − bi+1) + Snbn.

(2) AM-GM (Arithmetic Mean-Geometric Mean) InequalityIf a1, a2, ..., an are nonnegative real numbers, then

1n

n∑i=1

ai ≥ (a1a2...an)1n ,

with equality if and only if a1 = a2 = ... = an. This inequality is a specialcase of the Power Mean Inequality.

(3) Arithmetic Mean-Harmonic Mean (AM-HM) InequalityIf a1, a2, ..., an are positive real numbers, then

1n

n∑i=1

ai ≥1

1n

n∑i=1

1ai

with equality if and only if a1 = a2 = ... = an. This inequality is a specialcase of the Power Mean Inequality.

(4) Bernoulli’s InequalityFor any real numbers x > −1 and a > 1 we have (1 + x)a ≥ 1 + ax.

121

122 Solutions

(5) Cauchy-Schwarz’s InequalityFor any real numbers a1, a2, ..., an and b1, b2, ..., bn

(a21 + a2

2 + ... + a2n)(b2

1 + b22 + ... + b2

n) ≥

≥ (a1b1 + a2b2 + ... + anbn)2,

with equality if and only if ai and bi are proportional, i = 1, 2, ..., n.

(6) Cauchy-Schwarz’s Inequality for integralsIf a, b are real numbers, a < b, and f, g : [a, b] → R are integrable functions,then (∫ b

a

f(x)g(x)dx

)2

≤

(∫ b

a

f2(x)dx

)·

(∫ b

a

g2(x)dx

).

(7) Chebyshev’s InequalityIf a1 ≤ a2 ≤ . . . ≤ an and b1, b2, . . . , bn are real numbers, then

1)If b1 ≤ b2 ≤ . . . ≤ bn thenn∑

i=1

aibi ≥1n

(n∑

i=1

ai

)·

(n∑

i=1

bi

);

2)If b1 ≥ b2 ≥ . . . ≥ bn thenn∑

i=1

aibi ≤1n

(n∑

i=1

ai

)·

(n∑

i=1

bi

);

(8) Chebyshev’s Inequality for integralsIf a, b are real numbers, a < b, and f, g : [a, b] → R are integrable functionsand having the same monotonicity, then

(b− a)∫ b

a

f(x)g(x)dx ≥∫ b

a

f(x)dx ·∫ b

a

g(x)dx

and if one is increasing, while the other is decreasing the reversed inequalityis true.

(9) Convex functionA real-valued function f defined on an interval I of real numbers is convexif, for any x, y in I and any nonnegative numbers α, β with sum 1,

f(αx + βy) ≤ αf(x) + βf(y).

(10) ConvexityA function f(x) is concave up (down) on [a, b] ⊆ R if f(x) lies under(over) the line connecting (a1, f(a1)) and (b1, f(b1)) for all

a ≤ a1 < x < b1 ≤ b.

A function g(x) is concave up (down) on the Euclidean plane if it is concaveup (down) on each line in the plane, where we identify the line naturally


with R.Concave up and down functions are also called convex and concave, re-spectively.If f is concave up on an interval [a, b] and λ1, λ2, ..., λn are nonnegativenumbers wit sum equal to 1, then

λ1f(x1) + λ2f(x2) + ... + λnf(xn) ≥ f(λ1x1 + λ2x2 + ... + λnxn)

for any x1, x2, ..., xn in the interval [a, b]. If the function is concave down,the inequality is reversed. This is Jensen’s Inequality.

(11) Cyclic SumLet n be a positive integer. Given a function f of n variables, define thecyclic sum of variables (x1, x2, ..., xn) as∑

cyc

f(x1, x3, ..., xn) = f(x1, x2, ..., xn) + f(x2, x3, ..., xn, x1)

+... + f(an, a1, a2, ..., an−1).

(12) Holder’s Inequality

If r, s are positive real numbers such that1r

+1s

= 1, then for any positivereal numbers a1, a2, . . . , an,

b1, b2, . . . , bn,

n∑i=1

aibi

n≤

n∑

i=1

ari

n

1r

·

n∑

i=1

bsi

n

1s

.

(13) Huygens InequalityIf p1, p2, . . . , pn, a1, a2, . . . , an, b1, b2, . . . , bn are positive real numbers withp1 + p2 + . . . + pn = 1, then

n∏i=1

(ai + bi)pi ≥n∏

i=1

api

i +n∏

i=1

bpi

i .

(14) Mac Laurin’s InequalityFor any positive real numbers x1, x2, . . . , xn,

S1 ≥ S2 ≥ . . . ≥ Sn,

124 Solutions

where

Sk = k

√√√√√√√∑

1≤i1<i2<...<ik≤n

xi1 · xi2 · . . . · xik(n

k

) .

(15) Minkowski’s InequalityFor any real number r ≥ 1 and any positive real numbersa1, a2, . . . , an, b1, b2, . . . , bn,(

n∑i=1

(ai + bi)r

) 1r

≤

(n∑

i=1

ari

) 1r

+

(n∑

i=1

bri

) 1r

.

(16) Power Mean InequalityFor any positive real numbers a1, a2, . . . , an with sum equal to 1, andany positive real numbers x1, x2, . . . , xn, we define Mr = (a1x

r1 + a2x

r2 +

. . . + anxrn)

1r if r is a non-zero real and M0 = xa1

1 xa22 . . . xan

n , M∞ =max{x1, x2, . . . , xn}, M−∞ = min{x1, x2, . . . , xn}. Then for any reals s ≤ t

we have M−∞ ≤ Ms ≤ Mt ≤ M∞.

(17) Root Mean Square InequalityIf a1, a2, ..., an are nonnegative real numbers, then√

a21 + a2

2 + ... + a2n

n≥ a1 + a2 + ... + an

n,

with equality if and only if a1 = a2 = ... = an.

(18) Schur’s InequalityFor any positive real numbers x, y, z and any r > 0, xr(x−y)(x−z)+yr(y−z)(y−x)+ zr(z−x)(z− y) ≥ 0. The most common case is r = 1, which hasthe following equivalent forms:

1) x3 + y3 + z3 + 3xyz ≥ xy(x + y) + yz(y + z) + zx(z + x);

2) xyz ≥ (x + y − z)(y + z − z)(z + x− y);

3) if x + y + z = 1 then xy + yz + zx ≤ 1 + 9xyz

4.

(19) Suranyi’s InequalityFor any nonnegative real numbers a1, a2, . . . , an,

(n− 1)n∑

k=1

ank + n

n∏k=1

ak ≥

(n∑

k=1

ak

)·

(n∑

k=1

an−1k

).


(20) Turkevici’s InequalityFor any positive real numbers x, y, z, t,


(21) Weighted AM-GM InequalityFor any nonnegative real numbers a1, a2, ..., an, if w1, w2, ..., wn are nonneg-ative real numbers (weights) with sum 1, then

w1a1 + w2a2 + ... + wnan ≥ aw11 aw2

2 . . . awnn ,

with equality if and only if a1 = a2 = ... = an.

Further reading

1. Andreescu, T.; Feng, Z., 101 Problems in Algebra from the Training of the USAIMO Team, Australian Mathematics Trust, 2001.2. Andreescu, T.; Feng, Z., USA and International Mathematical Olympiads 2000,

2001, 2002, 2003, MAA.3. Andreescu, T.; Feng, Z., Mathematical Olympiads: Problems and Solutions fromaround the World, 1998-1999, 1999-2000, MAA.4. Andreescu, T.; Kedlaya, K., Mathematical Contests 1995-1996, 1996-1997,1997-1998: Olympiad Problems from around the World, with Solutions, AMC.5. Andreescu, T.; Enescu, B., Mathematical Olympiad Treasures, Birkhauser, 2003.6. Andreescu, T.; Gelca, R., Mathematical Olympiad Challenges, Birkhauser, 2000.7. Andreescu, T.; Andrica, D., 360 Problems for Mathematical Contests, GILPublishing House, 2002.8. Becheanu, M., Enescu, B., Inegalitati elementare. . . si mai putin elementare, GILPublishing House, 20029. Beckenbach, E., Bellman, R. Inequalities, Springer Verlag, Berlin, 196110. Drimbe, M.O., Inegalitati idei si metode, GIL Publishing House, 200311. Engel, A., Problem-Solving Strategies, Problem Books in Mathematics, Springer,1998.12. G.H. Littlewood, J.E. Polya, G., Inequalities, Cambridge University Press, 196713. Klamkin, M., International Mathematical Olympiads, 1978-1985, NewMathematical Library, Vol. 31, MAA, 1986.14. Larson, L. C., Problem-Solving Through Problems, Springer-Verlag, 1983.15. Lascu, M., Inegalitati, GIL Publishing House, 199416. Liu, A., Hungarian Problem Book III, New Mathematical Library, Vol. 42,MAA, 2001.17. Lozansky, E; Rousseau, C., Winning Solutions, Springer, 1996.18. Mitrinovic, D.S., Analytic inequalities, Springer Verlag, 197019. Panaitopol, L., Bandila, V., Lascu, M., Inegalitati, GIL Publishing House, 199520. Savchev, S.; Andreescu, T., Mathematical Miniatures, Anneli Lax NewMathematical Library, Vol. 43, MAA.

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molympiad.files.wordpress.com · Preface This work blends together classic inequality results with brand new problems, some of which devised only a few days ago. What could be special