The Equal Variable Method by Vasile Cirtoaje

8/14/2019 The Equal Variable Method by Vasile Cirtoaje

1/21

Volume 8 (2007), Issue 1, Article 15, 21 pp.

THE EQUAL VARIABLE METHOD

VASILE CRTOAJE

DEPARTMENT OF AUTOMATION AND COMPUTERSUNIVERSITY OF PLOIESTIBUCUREST I 39, ROMANIA

[email protected]

Received 01 March, 2006; accepted 17 April, 2006

Communicated by P.S. Bullen

ABSTRACT. The Equal Variable Method (called also n1 Equal Variable Method on the Math-links Site - Inequalities Forum) can be used to prove some difficult symmetric inequalities in-volving either three power means or, more general, two power means and an expression of formf(x1) + f(x2) + + f(xn).

Key words and phrases: Symmetric inequalities, Power means, EV-Theorem.

2000 Mathematics Subject Classification. 26D10, 26D20.

1. STATEMENT OF RESULTS

In order to state and prove the Equal Variable Theorem (EV-Theorem) we require the follow-ing lemma and proposition.

Lemma 1.1. Let a,b,c be fixed non-negative real numbers, not all equal and at most one of

them equal to zero, and let x y z be non-negative real numbers such thatx + y + z = a + b + c, xp + yp + zp = ap + bp + cp,

where p (, 0] (1, ). Forp = 0, the second equation is xyz = abc > 0. Then, thereexist two non-negative real numbers x1 andx2 with x1 < x2 such thatx [x1, x2]. Moreover,

(1) ifx = x1 andp 0, then 0 < x < y = z;(2) ifx = x1 andp > 1, then either0 = x < y z or0 < x < y = z;(3) ifx (x1, x2), then x < y < z;(4) ifx = x2, then x = y < z.

Proposition 1.2. Let a,b,c be fixed non-negative real numbers, not all equal and at most one

of them equal to zero, and let0 x y z such thatx + y + z = a + b + c, xp + yp + zp = ap + bp + cp,

059-06

upload by wWw.chuyenhungvuong.net
http://www.ams.org/msc/mailto:[email protected]


2/21

2 VASILE CRTOAJE

where p (, 0] (1, ). Forp = 0, the second equation is xyz = abc > 0. Letf(u) be adifferentiable function on (0, ), such thatg(x) = f

x

1p1

is strictly convex on (0, ), and

let

F3(x,y,z) = f(x) + f(y) + f(z).

(1) If p 0 , then F3 is maximal only for 0 < x = y < z , and is minimal only for0 < x < y = z;

(2) Ifp > 1 and eitherf(u) is continuous atu = 0 or limu0

f(u) = , then F3 is maximalonly for0 < x = y < z, and is minimal only for eitherx = 0 or0 < x < y = z.

Theorem 1.3 (Equal Variable Theorem (EV-Theorem)). Let a1, a2, . . . , an (n 3) be fixednon-negative real numbers, and let 0 x1 x2 xn such that

x1 + x2 + + xn = a1 + a2 + + an,xp1 + x

p2 +

+ xpn = a

p1 + a

p2 +

+ apn,

where p is a real number, p = 1. Forp = 0, the second equation is x1x2 xn = a1a2 an >0. Letf(u) be a differentiable function on (0, ) such that

g(x) = f

x1

p1

is strictly convex on (0, ), and letFn(x1, x2, . . . , xn) = f(x1) + f(x2) + + f(xn).

(1) If p 0 , then Fn is maximal for 0 < x1 = x2 = = xn1 xn, and is minimal for0 < x1

x2 = x3 =

= xn;

(2) If p > 0 and either f(u) is continuous at u = 0 or limu0 f(u) = , then Fn ismaximal for 0 x1 = x2 = = xn1 xn, and is minimal for either x1 = 0 or0 < x1 x2 = x3 = = xn.

Remark 1.4. Let 0 < < . If the function f is differentiable on (, ) and the function

g(x) = f

x1

p1

is strictly convex on (p1, p1) or (p1, p1), then the EV-Theorem

holds true for x1, x2, . . . , xn (, ).By Theorem 1.3, we easily obtain some particular results, which are very useful in applica-

tions.

Corollary 1.5. Let a1, a2, . . . , an (n 3) be fixed non-negative numbers, and let 0 x1 x2 xn such that

x1 + x2 + + xn = a1 + a2 + + an,x21 + x

22 + + x2n = a21 + a22 + + a2n.

Letf be a differentiable function on (0, ) such thatg(x) = f(x) is strictly convex on (0, ).Moreover, eitherf(x) is continuous atx = 0 or lim

x0f(x) = . Then,

Fn = f(x1) + f(x2) + + f(xn)

is maximal for 0 x1 = x2 = = xn1 xn , and is minimal for either x1 = 0 or0 < x1 x2 = x3 = = xn.

J. Inequal. Pure and Appl. Math., 8(1) (2007), Art. 15, 21 pp. http://jipam.vu.edu.au/
http://jipam.vu.edu.au/


3/21

EQUAL VARIABLE METHOD 3

Corollary 1.6. Let a1, a2, . . . , an (n 3) be fixed positive numbers, and let 0 < x1 x2 xn such that

x1 + x2 + + xn = a1 + a2 + + an,1

x1 +

1

x2 + +1

xn =

1

a1 +

1

a2 + +1

an .

Let f be a differentiable function on (0, ) such that g(x) = f

1x

is strictly convex on

(0, ). Then,Fn = f(x1) + f(x2) + + f(xn)

is maximal for0 < x1 = x2 = = xn1 xn, and is minimal for0 < x1 x2 = x3 = =xn.

Corollary 1.7. Let a1, a2, . . . , an (n 3) be fixed positive numbers, and let 0 < x1 x2 xn such that

x1 + x2 + + xn = a1 + a2 + + an, x1x2 xn = a1a1 an.Letf be a differentiable function on (0, ) such thatg(x) = f 1

x

is strictly convex on (0, ).

Then,

Fn = f(x1) + f(x2) + + f(xn)is maximal for0 < x1 = x2 = = xn1 xn, and is minimal for0 < x1 x2 = x3 = =xn.

Corollary 1.8. Let a1, a2, . . . , an (n 3) be fixed non-negative numbers, and let 0 x1 x2 xn such that

x1 + x2 +

+ xn = a1 + a2 +

+ an,

xp1 + xp2 + + xpn = ap1 + ap2 + + apn,where p is a real number, p = 0 andp = 1.

(a) Forp < 0, P = x1x2 xn is minimal when 0 < x1 = x2 = = xn1 xn, and ismaximal when 0 < x1 x2 = x3 = = xn.

(b) Forp > 0, P = x1x2 xn is maximal when 0 x1 = x2 = = xn1 xn, and isminimal when eitherx1 = 0 or0 < x1 x2 = x3 = = xn.

Corollary 1.9. Let a1, a2, . . . , an (n 3) be fixed non-negative numbers, let 0 x1 x2 xn such that

x1 + x2 +

+ xn = a1 + a2 +

+ an,

xp1 + x

p2 + + xpn = ap1 + ap2 + + apn,

and letE = xq1 + xq2 + + xqn.

Case 1. p 0 (p = 0 yields x1x2 xn = a1a2 an > 0).(a) For q (p, 0) (1, ), E is maximal when 0 < x1 = x2 = = xn1 xn, and is

minimal when 0 < x1 x2 = x3 = = xn.(b) For q (, p) (0, 1), E is minimal when 0 < x1 = x2 = = xn1 xn, and is

maximal when 0 < x1 x2 = x3 = = xn.Case 2. 0 < p < 1.

(a) For q (0, p) (1, ), E is maximal when 0 x1 = x2 = = xn1 xn, and isminimal when either x1 = 0 or 0 < x1 x2 = x3 = = xn.



4/21

4 VASILE CRTOAJE

(b) For q (, 0) (p, 1), E is minimal when 0 x1 = x2 = = xn1 xn, and ismaximal when either x1 = 0 or 0 < x1 x2 = x3 = = xn.

Case 3. p > 1.

(a) For q

(0, 1)

(p,

), E is maximal when 0

x1 = x2 =

= xn1

xn, and is

minimal when either x1 = 0 or 0 < x1 x2 = x3 = = xn.(b) For q (, 0) (1, p), E is minimal when 0 x1 = x2 = = xn1 xn, and is

maximal when either x1 = 0 or 0 < x1 x2 = x3 = = xn.

2. PROOFS

Proof of Lemma 1.1. Let a b c. Note that in the excluded cases a = b = c and a = b = 0,there is a single triple (x,y,z) which verifies the conditions

x + y + z = a + b + c and xp + yp + zp = ap + bp + cp.

Consider now three cases: p = 0, p < 0 and p > 1.A. Case p = 0 (xyz = abc > 0). Let S = a+b+c

3and P = 3

abc, where S > P > 0 by AM-GM

Inequality. We have

x + y + z = 3S, xyz = P3,

and from 0 < x y z and x < z, it follows that 0 < x < P. Now letf = y + z 2yz.

It is clear that f 0, with equality if and only ify = z. Writing f as a function ofx,

f(x) = 3S

x

2PPx

,

we have

f(x) =P

x

P

x 1 > 0,

and hence the function f(x) is strictly increasing. Since f(P) = 3(S P) > 0, the equationf(x) = 0 has a unique positive root x1, 0 < x1 < P. From f(x) 0, it follows that x x1.Sub-case x = x1. Since f(x) = f(x1) = 0 and f = 0 implies y = z, we have 0 < x < y = z.

Sub-case x > x1. We have f(x) > 0 and y < z. Consider now that y and z depend on x. Fromx + y(x) + z(x) = 3S and x y(x) z(x) = P3, we get 1 + y + z = 0 and 1

x+ y

y+ z

z= 0.

Hence,y(x) =

y(x z)x(z y) , z

(x) =z(y x)x(z y) .

Since y(x) < 0, the function y(x) is strictly decreasing. Since y(x1) > x1 (see sub-casex = x1), there exists x2 > x1 such that y(x2) = x2, y(x) > x for x1 < x < x2 and y(x) < xfor x > x2. Taking into account that y x, it follows that x1 < x x2. On the other hand, wesee that z(x) > 0 for x1 < x < x2. Consequently, the function z(x) is strictly increasing, andhence z(x) > z(x1) = y(x1) > y(x). Finally, we conclude that x < y < z for x (x1, x2),and x = y < z for x = x2.

B. Case p < 0. Denote S = a+b+c3

and R = ap+bp+cp3

1p . Taking into account thatx + y + z = 3S, xp + yp + zp = 3Rp,



5/21


from 0 < x y z and x < z we get x < S and 3 1p R < x < R. Let

h = (y + z)

yp + zp

2

1p

2.

By the AM-GM Inequality, we have

h 2yz 1yz

2 = 0,

with equality if and only ify = z. Writing now h as a function ofx,

h(x) = (3S x)

3Rp xp2

1p

2,

from

h(x) =3Rp

2 3Rp xp

2 1pp

S

xR

x p

1 > 0it follows that h(x) is strictly increasing. Since h(x) 0 and h

31p R

= 2, the equationh(x) = 0 has a unique root x1 and x x1 > 3

1p R.

Sub-case x = x1. Since f(x) = f(x1) = 0, and f = 0 implies y = z, we have 0 < x < y = z.

Sub-case x > x1. We have h(x) > 0 and y < z. Consider now that y and z depend on x.From x + y(x) + z(x) = 3S and xp + y(x)p + z(x)p = 3Rp, we get 1 + y + z = 0 andxp1 + yp1y + zp1z = 0, and hence

y(x) =xp1 zp1zp1

yp1

, z(x) =xp1 yp1yp1

zp1

.

Since y(x) > 0, the function y(x) is strictly decreasing. Since y(x1) > x1 (see sub-casex = x1), there exists x2 > x1 such that y(x2) = x2, y(x) > x for x1 < x < x2, and y(x) < x forx > x2. The condition y x yields x1 < x x2. We see now that z(x) > 0 for x1 < x < x2.Consequently, the function z(x) is strictly increasing, and hence z(x) > z(x1) = y(x1) > y(x).Finally, we have x < y < z for x (x1, x2) and x = y < z for x = x2.C. Case p > 1. Denoting S = a+b+c

3and R =

ap+bp+cp

3

1p yields

x + y + z = 3S, xp + yp + zp = 3Rp.

By Jensens inequality applied to the convex function g(u) = up, we have R > S, and hence

x < S < R. Leth =

2

y + z

yp + zp

2

1p

1.By Jensens Inequality, we get h 0, with equality if only ify = z. From

h(x) =2

3S x

3Rp xp2

1p

1

and

h(x) =3

(3S x)2

3Rp xp2

1pp

(Rp Sxp1) > 0,

it follows that the function h(x) is strictly increasing, and h(x) 0 implies x x1. In the caseh(0) 0 we have x1 = 0, and in the case h(0) < 0 we have x1 > 0 and h(x1) = 0.



6/21

6 VASILE CRTOAJE

Sub-case x = x1. Ifh(0) 0, then 0 = x1 < y(x1) z(x1). Ifh(0) < 0, then h(x1) = 0, andsince h = 0 implies y = z, we have 0 < x1 < y(x1) = z(x1).

Sub-case x > x1. Since h(x) is strictly increasing, for x > x1 we have h(x) > h(x1) 0,hence h(x) > 0 and y < z. From x + y(x) + z(x) = 3S and xp + yp(x) + zp(x) = 3Rp, we get

y(x) = xp1 zp1

zp1 yp1 , z(x) = y

p1 xp1zp1 yp1 .

Since y(x) < 0, the function y(x) is strictly decreasing. Taking account of y(x1) > x1 (seesub-case x = x1), there exists x2 > x1 such that y(x2) = x2, y(x) > x for x1 < x < x2,and y(x) < x for x > x2. The condition y x implies x1 < x x2. We see now thatz(x) > 0 for x1 < x < x2. Consequently, the function z(x) is strictly increasing, and hencez(x) > z(x1) y(x1) > y(x). Finally, we conclude that x < y < z for x (x1, x2), andx = y < z for x = x2.

Proof of Proposition 1.2. Consider the function

F(x) = f(x) + f(y(x)) + f(z(x))defined on x [x1, x2]. We claim that F(x) is minimal for x = x1 and is maximal for x = x2.If this assertion is true, then by Lemma 1.1 it follows that:

(a) F(x) is minimal for 0 < x = y < z in the case p 0, or for either x = 0 or0 < x < y = z in the case p > 1;

(b) F(x) is maximal for 0 < x = y < z.

In order to prove the claim, assume that x (x1, x2). By Lemma 1.1, we have 0 < x < y 0.

These results imply F(x) > 0. Consequently, the function F(x) is strictly increasing forx (x1, x2). Excepting the trivial case when p > 1, x1 = 0 and limu0

f(u) = , the function



7/21


8/21

8 VASILE CRTOAJE

and

g(x) = f

x1

p1

= q2(q 1)(q p)x q1p1 ,

g(x) =q2(q 1)2(q p)2

(p 1)2

x2p11p .

Since g(x) > 0 for x > 0, the function g(x) is strictly convex on (0, ), and the conclusionfollows by Theorem 1.3.

3. APPLICATIONS

Proposition 3.1. Letx,y,z be non-negative real numbers such thatx+y+z = 2. Ifr0 r 3,where r0 =

ln 2ln 3ln 2 1.71, then

xr(y + z) + yr(z + x) + zr(x + y) 2.Proof. Rewrite the inequality in the homogeneous form

xr+1 + yr+1 + zr+1 + 2

x + y + z2

r+1 (x + y + z)(xr + yr + zr),and apply Corollary 1.9 (case p = r and q = r + 1):

If0 x y z such thatx + y + z = constant and

xr + yr + zr = constant,

then the sum xr+1 + yr+1 + zr+1 is minimal when either x = 0 or 0 < x y = z.Case x = 0. The initial inequality becomes

yz(yr1 + zr1)

2,

where y + z = 2. Since 0 < r 1 2, by the Power Mean inequality we haveyr1 + zr1

2

y2 + z2

2

r12

.

Thus, it suffices to show that

yz

y2 + z2

2

r12

1.Taking account of

y2 + z2

2=

2(y2 + z2)

(y + z)2 1 and r 1

2 1,

we have

1 yz

y2 + z2

2

r12

1 yz

y2 + z2

2

=(y + z)4

16 yz(y

2 + z2)

2

=(y z)4

16 0.

Case 0 < x y = z. In the homogeneous inequality we may leave aside the constraintx + y + z = 2, and consider y = z = 1, 0 < x

1. The inequality reduces to

1 + x2

r+1 xr x 1 0.



9/21


Since

1 + x2

r+1is increasing and xr is decreasing in respect to r, it suffices to consider r = r0.

Let

f(x) =

1 +x

2

r0+1 xr0 x 1.We have

f(x) =r0 + 1

2

1 +

x

2

r0 r0xr01 1,1

r0f(x) =

r0 + 1

4

1 +

x

2

r0 r0 1x2r0

.

Since f(x) is strictly increasing on (0, 1], f(0+) = and1

r0f(1) =

r0 + 1

4

3

2

r0 r0 + 1

=r0 + 1

2 r0 + 1 =

3 r02

> 0,

there exists x1 (0, 1) such that f(x1) = 0, f(x) < 0 for x (0, x1), and f(x) > 0 forx (x1, 1]. Therefore, the function f(x) is strictly decreasing for x [0, x1], and strictlyincreasing for x [x1, 1]. Since

f(0) =r0 1

2> 0 and f(1) =

r0 + 1

2

3

2

r0 2

= 0,

there exists x2 (0, x1) such that f(x2) = 0, f(x) > 0 for x [0, x2), and f(x) < 0 forx (x2, 1). Thus, the function f(x) is strictly increasing for x [0, x2], and strictly decreasingfor x [x2, 1]. Since f(0) = f(1) = 0, it follows that f(x) 0 for 0 < x 1, establishing thedesired result.

For x y z, equality occurs when x = 0 and y = z = 1. Moreover, for r = r0, equalityholds again when x = y = z = 1.

Proposition 3.2 ([12]). Letx,y,z be non-negative real numbers such that xy + yz + zx = 3.If1 < r 2, then

xr(y + z) + yr(z + x) + zr(x + y) 6.Proof. Rewrite the inequality in the homogeneous form

xr(y + z) + yr(z + x) + zr(x + y) 6

xy + yz + zx

3

r+12

.

For convenience, we may leave aside the constraint xy + yz + zx = 3. Using now the constraintx + y + z = 1, the inequality becomes

xr(1 x) + yr(1 y) + zr(1 z) 6

1 x2 y2 z26

r+12

.

To prove it, we will apply Corollary 1.5 to the function f(u) = ur(1 u) for 0 u 1. Wehave f(u) = rur1 + (r + 1)ur and

g(x) = f(x) = rxr1 + (r + 1)xr, g(x) = r(r 1)xr3[(r + 1)x + 2 r].Since g(x) > 0 for x > 0, g(x) is strictly convex on [0,

). According to Corollary 1.5,

if 0 x y z such that x + y + z = 1 and x2 + y2 + z2 = constant, then the sumf(x) + f(y) + f(z) is maximal for 0 x = y z.



10/21

10 VASILE CRTOAJE

Thus, we have only to prove the original inequality in the case x = y z. This means, toprove that 0 < x 1 y and x2 + 2xz = 3 implies

xr(x + z) + xzr 3.

Let f(x) = xr

(x + z) + xzr

3, with z =3x2

2x .Differentiating the equation x2 + 2xz = 3 yields z = (x+z)x

. Then,

f(x) = (r + 1)xr + rxr1z + zr + (xr + rxzr1)z

= (xr1 zr1)[rx + (r 1)z] 0.The function f(x) is strictly decreasing on [0, 1], and hence f(x) f(1) = 0 for 0 < x 1.Equality occurs if and only ifx = y = z = 1.

Proposition 3.3 ([5]). Ifx1, x2, . . . , xn are positive real numbers such that

x1 + x2 +

+ xn =1

x1+

1

x2+

+

1

xn,

then1

1 + (n 1)x1 +1

1 + (n 1)x2 + +1

1 + (n 1)xn 1.

Proof. We have to consider two cases.

Case n = 2. The inequality is verified as equality.

Case n 3. Assume that 0 < x1 x2 xn, and then apply Corollary 1.6 to the functionf(u) = 1

1+(n1)u for u > 0. We have f(u) = (n1)

[1+(n1)u]2 and

g(x) = f 1x = (n 1)x(x + n 1)2 ,

g(x) =3(n 1)2

2

x (

x + n 1)4 .

Since g(x) > 0, g(x) is strictly convex on (0, ). According to Corollary 1.6, if0 < x1 x2 xn such that

x1 + x2 + + xn = constant and1

x1+

1

x2+ + 1

xn= constant,

then the sum f(x1) + f(x2) + + f(xn) is minimal when 0 < x1 x2 = x3 = = xn.Thus, we have to prove the inequality

1

1 + (n 1)x +n 1

1 + (n 1)y 1,

under the constraints 0 < x 1 y and

x + (n 1)y = 1x

+n 1

y.

The last constraint is equivalent to

(n 1)(y 1) = y(1 x2)x(1 + y)

.



11/21


Since1

1 + (n 1)x +n 1

1 + (n 1)y 1

=1

1 + (n 1)x 1

n+

n 11 + (n 1)y

n 1n

=(n 1)(1 x)n[1 + (n 1)x]

(n 1)2(y 1)n[1 + (n 1)y]

=(n 1)(1 x)n[1 + (n 1)x]

(n 1)y(1 x2)nx(1 + y)[1 + (n 1)y] ,

we must show that

x(1 + y)[1 + (n 1)y] y(1 + x)[1 + (n 1)x],which reduces to

(y

x)[(n

1)xy

1]

0.

Since y x 0, we have still to prove that(n 1)xy 1.

Indeed, from x + (n 1)y = 1x

+ n1y

we get xy = y+(n1)xx+(n1)y , and hence

(n 1)xy 1 = n(n 2)xx + (n 1)y > 0.

For n 3, one has equality if and only ifx1 = x2 = = xn = 1. Proposition 3.4 ([10]). Leta1, a2, . . . , an be positive real numbers such that a1a2 an = 1. Ifm is a positive integer satisfying m

n

1, then

am1 + am2 + + amn + (m 1)n m

1

a1+

1

a2+ + 1

an

.

Proof. For n = 2 (hence m 1), the inequality reduces toam1 + a

m2 + 2m 2 m(a1 + a2).

We can prove it by summing the inequalities am1 1+ m(a11) and am2 1+ m(a21), whichare straightforward consequences of Bernoullis inequality. For n 3, replacing a1, a2, . . . , anby 1

x1, 1x2

, . . . , 1xn

, respectively, we have to show that

1

xm1

+1

xm2

+

+

1

xmn+ (m

1)n

m(x1 + x2 +

+ xn)

for x1x2 xn = 1. Assume 0 < x1 x2 xn and apply Corollary 1.9 (case p = 0 andq = m):

If0 < x1 x2 xn such thatx1 + x2 + + xn = constant and

x1x2 xn = 1,then the sum 1

xm1

+ 1xm2

+ + 1xmn

is minimal when 0 < x1 = x2 = = xn1 xn.Thus, it suffices to prove the inequality for x1 = x2 = = xn1 = x 1, xn = y and

xn1y = 1, when it reduces to:

n 1xm

+ 1ym

+ (m 1)n m(n 1)x + my.



12/21

12 VASILE CRTOAJE

By the AM-GM inequality, we have

n 1xm

+ (m n + 1) mxn1

= my.

Then, we have still to show that

1

ym 1 m(n 1)(x 1).

This inequality is equivalent to

xmnm 1 m(n 1)(x 1) 0and

(x 1)[(xmnm1 1) + (xmnm2 1) + + (x 1)] 0.The last inequality is clearly true. For n = 2 and m = 1, the inequality becomes equality.Otherwise, equality occurs if and only ifa1 = a2 = = an = 1.

Proposition 3.5 ([6]). Letx1, x2, . . . , xn be non-negative real numbers such thatx1+x2+ +xn = n. Ifk is a positive integer satisfying 2 k n + 2, andr =

n

n1k1 1, then

xk1 + xk2 + + xkn n nr(1 x1x2 xn).

Proof. Ifn = 2, then the inequality reduces to xk1 + xk2 2 (2k 2)x1x2. For k = 2 and

k = 3, this inequality becomes equality, while for k = 4 it reduces to 6x1x2(1 x1x2) 0,which is clearly true.

Consider now n 3 and 0 x1 x2 xn. Towards proving the inequality,we will apply Corollary 1.8 (case p = k > 0): If0 x1 x2 xn such thatx1 + x2 + + xn = n and xk1 + xk2 + + xkn = constant, then the product x1x2 xn isminimal when either x1 = 0 or 0 < x1 x2 = x3 = = xn.Case x1 = 0. The inequality reduces to

xk2 + + xkn nk

(n 1)k1 ,

with x2 + + xn = n, This inequality follows by applying Jensens inequality to the convexfunction f(u) = uk:

xk2 + + xkn (n 1)

x2 + + xnn 1

k.

Case 0 < x1

x2 = x3 =

= xn. Denoting x1 = x and x2 = x3 =

= xn = y, we have

to prove that for 0 < x 1 y and x + (n 1)y = n, the inequality holds:xk + (n 1)yk + nrxyn1 n(r + 1) 0.

Write the inequality as f(x) 0, wheref(x) = xk + (n 1)yk + nrxyn1 n(r + 1), with y = n x

n 1 .

We see that f(0) = f(1) = 0. Since y = 1n1 , we have

f(x) = k(xk1 yk1) + nryn2(y x)= (y

x)[nryn2

k(yk2 + yk3x +

+ xk2)]

= (y x)yn2[nr kg(x)],



13/21


where

g(x) =1

ynk+

x

ynk+1+ + x

k2

yn2.

Since the function y(x) = nxn1

is strictly decreasing, the function g(x) is strictly increasing for

2 k n. For k = n + 1, we haveg(x) = y + x +

x2

y+ + x

n1

yn2

=(n 2)x + n

n 1 +x2

y+ + x

n1

yn2,

and for k = n + 2, we have

g(x) = y2 + yx + x2 +x3

y+ + x

n

yn2

= (n2

3n + 3)x2

+ n(n 3)x + n2

(n 1)2 + x3

y+ + x

n

yn2.

Therefore, the function g(x) is strictly increasing for 2 k n + 2, and the functionh(x) = nr kg(x)

is strictly decreasing. Note that

f(x) = (y x)yn2h(x).We assert that h(0) > 0 and h(1) < 0. If our claim is true, then there exists x1 (0, 1) such thath(x1) = 0, h(x) > 0 for x

[0, x1), and h(x) < 0 for x

(x1, 1]. Consequently, f(x) is strictly

increasing for x [0, x1], and strictly decreasing for x [x1, 1]. Since f(0) = f(1) = 0, itfollows that f(x) 0 for 0 < x 1, and the proof is completed.

In order to prove that h(0) > 0, we assume that h(0) 0. Then, h(x) < 0 for x (0, 1),f(x) < 0 for x (0, 1), and f(x) is strictly decreasing for x [0, 1], which contradictsf(0) = f(1). Also, ifh(1) 0, then h(x) > 0 for x (0, 1), f(x) > 0 for x (0, 1), andf(x) is strictly increasing for x [0, 1], which also contradicts f(0) = f(1).

For n 3 and x1 x2 xn, equality occurs when x1 = x2 = = xn = 1, and alsowhen x1 = 0 and x2 = = xn = nn1 . Remark 3.6. For k = 2, k = 3 and k = 4, we get the following nice inequalities:

(n 1)(x2

1 + x2

2 + + x2

n) + nx1x2 xn n2

,

(n 1)2(x31 + x32 + + x3n) + n(2n 1)x1x2 xn n3,(n 1)3(x41 + x42 + + x4n) + n(3n2 3n + 1)x1x2 xn n4.

Remark 3.7. The inequality for k = n was posted in 2004 on the Mathlinks Site - InequalitiesForum by Gabriel Dospinescu and Calin Popa.

Proposition 3.8 ([11]). Letx1, x2, . . . , xn be positive real numbers such that 1x1 +1x2

+ + 1xn

=n. Then

x1 + x2 +

+ xn

n

en1(x1x2

xn

1),

where en1 =

1 + 1n1

n1< e.



14/21

14 VASILE CRTOAJE

Proof. Replacing each of the xi by 1ai , the statement becomes as follows:Ifa1, a2, . . . , an are positive numbers such that a1 + a2 + + an = n, then

a1a2 an1

a1+

1

a2+ + 1

an n + en1

en1.

It is easy to check that the inequality holds for n = 2. Consider now n 3, assume that0 < a1 a2 an and apply Corollary 1.8 (case p = 1): If0 < a1 a2 ansuch that a1 + a2 + + an = n and 1a1 + 1a2 + + 1an = constant, then the product a1a2 anis maximal when 0 < a1 a2 = a3 = = an.

Denoting a1 = x and a2 = a3 = = an = y, we have to prove that for 0 < x 1 y 0. If h(0)

0, then h(x) > 0 for x

(0, 1),

hence f(x) > 0 for x (0, 1), and f(x) is strictly increasing for x [0, 1], which contradictsf(0) = f(1). Also, h(1) = en1 2 > 0.

From h(0) < 0 and h(1) > 0, it follows that there exists x1 (0, 1) such that h(x1) = 0,h(x) < 0 for x [0, x1), and h(x) > 0 for x (x1, 1]. Consequently, f(x) is strictly decreasingfor x [0, x1], and strictly increasing for x [x1, 1]. Since f(0) = f(1) = 0, it follows thatf(x) 0 for 0 x 1.

For n 3, equality occurs when x1 = x2 = = xn = 1. Proposition 3.9 ([9]). Ifx1, x2, . . . , xn are positive real numbers, then

xn1 + xn2 + + xnn + n(n 1)x1x2 xn

x1x2 xn(x1 + x2 + + xn)

1x1

+ 1x2

+ + 1xn

.

Proof. For n = 2, one has equality. Assume now that n 3, 0 < x1 x2 xn andapply Corollary 1.9 (case p = 0): If0 < x1 x2 xn such that

x1 + x2 + + xn = constant andx1x2 xn = constant,

then the sum xn1 + xn2 + + xnn is minimal and the sum 1x1 + 1x2 + + 1xn is maximal when

0 < x1 x2 = x3 = = xn.Thus, it suffices to prove the inequality for 0 < x1

1 and x2 = x3 =

= xn = 1. The

inequality becomesxn1 + (n 2)x1 (n 1)x21,



15/21


and is equivalent to

x1(x1 1)[(xn21 1) + (xn31 1) + + (x1 1)] 0,which is clearly true. For n 3, equality occurs if and only ifx1 = x2 = = xn.

Proposition 3.10 ([14]). Ifx1, x2, . . . , xn are non-negative real numbers, then(n 1)(xn1 + xn2 + + xnn) + nx1x2 xn

(x1 + x2 + + xn)(xn11 + xn12 + + xn1n ).Proof. For n = 2, one has equality. For n 3, assume that 0 x1 x2 xnand apply Corollary 1.9 (case p = n and q = n 1) and Corollary 1.8 (case p = n): If0 x1 x2 xn such that

x1 + x2 + + xn = constant andxn1 + x

n2 + + xnn = constant,

then the sum xn1

1 + xn1

2 + + xn1

n is maximal and the product x1x2 xn is minimal wheneither x1 = 0 or 0 < x1 x2 = x3 = = xn.So, it suffices to consider the cases x1 = 0 and 0 < x1 x2 = x3 = = xn.

Case x1 = 0. The inequality reduces to

(n 1)(xn2 + + xnn) (x2 + + xn)(xn12 + + xn1n ),which immediately follows by Chebyshevs inequality.

Case 0 < x1 x2 = x3 = = xn. Setting x2 = x3 = = xn = 1, the inequality reducesto:

(n 2)xn1 + x1 (n 1)xn11 .Rewriting this inequality as

x1(x1 1)[xn31 (x1 1) + xn41 (x21 1) + + (xn21 1)] 0,we see that it is clearly true. For n 3 and x1 x2 xn equality occurs whenx1 = x2 = = xn, and for x1 = 0 and x2 = = xn. Proposition 3.11 ([8]). Ifx1, x2, . . . , xn are positive real numbers, then

(x1 + x2 + + xn n)

1

x1+

1

x2+ + 1

xn n

+ x1x2 xn + 1

x1x2 xn 2.

Proof. For n = 2, the inequality reduces to

(1

x1)2(1

x2)

2

x1x2 0.For n 3, assume that 0 < x1 x2 xn. Since the inequality preserves its formby replacing each number xi with 1xi , we may consider x1x2 xn 1. So, by the AM-GMinequality we get

x1 + x2 + + xn n n n

x1x2 xn n 0,and we may apply Corollary 1.9 (case p = 0 and q = 1): If0 x1 x2 xn suchthat

x1 + x2 + + xn = constant andx1x2

xn = constant,

then the sum 1x1

+ 1x2

+ + 1xn

is minimal when 0 < x1 = x2 = = xn1 xn.



16/21


17/21


We have y = (n 1) andf(x)n 1 =

1

x 1

y+

2

n 1(x y)(n 1)x2 + y2 =

(y x)(n 1x y)2xy[(n 1)x2 + y2] 0.

Therefore, the function f(x) is strictly increasing on (0, 1] and hence f(x) f(1) = 0. Equal-ity occurs if and only ifx1 = x2 = = xn = 1. Remark 3.13. For n = 5, we get the following nice statement:

Ifa,b,c,d,e are positive real numbers such that a2 + b2 + c2 + d2 + e2 = 5, then

abcde(a4 + b4 + c4 + d4 + e4) 5.Proposition 3.14 ([4]). Letx,y,z be non-negative real numbers such that xy + yz + zx = 3,and let

p ln 9 ln 4ln 3

0.738.Then,

xp + yp + zp 3.Proof. Let r = ln 9ln 4

ln 3. By the Power-Mean inequality, we have

xp + yp + zp

3

xr + yr + zr

3

pr

.

Thus, it suffices to show thatxr + yr + zr 3.

Let x y z. We consider two cases.Case x = 0. We have to show that yr + zr

3 for yz = 3. Indeed, by the AM-GM inequality,

we getyr + zr 2(yz)r/2 = 2 3r/2 = 3.

Case x > 0. The inequality xr + yr + zr 3 is equivalent to the homogeneous inequality

xr + yr + zr 3xyz

3

r2

1

x+

1

y+

1

z

r2

.

Setting x = a1r , y = b

1r , z = c

1r (0 < a b c), the inequality becomes

a + b + c 3abc

3 1

2

a1r + b

1r + c

1r

r

2

.

Towards proving this inequality, we apply Corollary 1.9 (case p = 0, q = 1r

): If0 < a b csuch that a + b + c = constant and abc = constant, then the sum a

1r + b

1r + c

1r is maximal

when 0 < a b = c.So, it suffices to prove the inequality for 0 < a b = c; that is, to prove the homogeneous

inequality in x,y,z for 0 < x y = z. Without loss of generality, we may leave aside theconstraint xy + yz + zx = 3, and consider y = z = 1 and 0 < x 1. The inequality reduces to

xr + 2 3

2x + 1

3

r2

.

Denoting

f(x) = ln xr + 2

3 r

2ln 2x + 1

3,



18/21

18 VASILE CRTOAJE

we have to show that f(x) 0 for 0 < x 1. The derivative

f(x) =rxr1

xr + 2 r

2x + 1=

r(x 2x1r + 1)x1r(xr + 2)(2x + 1)

has the same sign as g(x) = x 2x1r + 1. Since g(x) = 1 2(1r)xr , we see that g(x) < 0for x (0, x1), and g(x) > 0 for x (x1, 1], where x1 = (2 2r)1/r 0.416. The functiong(x) is strictly decreasing on [0, x1], and strictly increasing on [x1, 1]. Since g(0) = 1 andg(1) = 0, there exists x2 (0, 1) such that g(x2) = 0, g(x) > 0 for x [0, x2) and g(x) < 0for x (x2, 1). Consequently, the function f(x) is strictly increasing on [0, x2] and strictlydecreasing on [x2, 1]. Since f(0) = f(1) = 0, we have f(x) 0 for 0 < x 1, establishingthe desired result.

Equality occurs for x = y = z = 1. Additionally, for p = ln 9ln 4ln 3

and x y z, equalityholds again for x = 0 and y = z =

3.

Proposition 3.15 ([7]). Letx,y,z be non-negative real numbers such that x + y + z = 3, andletp ln 9ln 8ln 3ln 2 0.29. Then,

xp + yp + zp xy + yz + zx.Proof. For p 1, by Jensens inequality we have

xp + yp + zp 3

x + y + z

3

p

= 3 =1

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

Assume now p < 1. Let r = ln 9ln 8ln 3ln 2 and x y z. The inequality is equivalent to thehomogeneous inequality

2(xp + yp + zp)

x + y + z

3

2p+ x2 + y2 + z2 (x + y + z)2.

By Corollary 1.9 (case 0 < p < 1 and q = 2), ifx y z such that x + y + z = constantand xp + yp + zp = constant, then the sum x2 + y2 + z2 is minimal when either x = 0 or0 < x y = z.Case x = 0. Returning to our original inequality, we have to show that yp + zp yz fory + z = 3. Indeed, by the AM-GM inequality, we get

yp + zp yz 2(yz)p2 yz= (yz)

p

2 [2 (yz) 2p2 ]

(yz) p2

2

y + z

2

2p

= (yz)p

2

2

3

2

2p

(yz)p

2

2 322r

= 0.



19/21


Case 0 < x y = z. In the homogeneous inequality, we may leave aside the constraintx + y + z = 3, and consider y = z = 1 and 0 < x 1. Thus, the inequality reduces to

(xp + 2)x + 2

3 2p

2x + 1.

To prove this inequality, we consider the function

f(x) = ln(xp + 2) + (2 p) ln x + 23

ln(2x + 1).We have to show that f(x) 0 for 0 < x 1 and r p < 1. We have

f(x) =pxp1

xp + 2+

2 px + 2

22x + 1

=2g(x)

x1p(xp + 2)(2x + 1),

whereg(x) = x2 + (2p 1)x + p + 2(1 p)x2p (p + 2)x1p,

and g(x) = 2x + 2p 1 + 2(1 p)(2 p)x1p (p + 2)(1 p)xp,g(x) = 2 + 2(1 p)2(2 p)xp + p(p + 2)(1 p)xp1.

Since g(x) > 0, the first derivative g(x) is strictly increasing on (0, 1]. Taking into accountthat g(0+) = and g(1) = 3(1 p) + 3p2 > 0, there is x1 (0, 1) such that g(x1) = 0,g(x) < 0 for x (0, x1)and g(x) > 0 for x (x1, 1]. Therefore, the function g(x) is strictlydecreasing on [0, x1] and strictly increasing on [x1, 1]. Since g(0) = p > 0 and g(1) = 0, thereis x2 (0, x1) such that g(x2) = 0, g(x) > 0 for x [0, x2) and g(x) < 0 for x (x2, 1]. Wehave also f(x2) = 0, f(x) > 0 for x (0, x2) and f(x) < 0 for x (x2, 1]. According to thisresult, the function f(x) is strictly increasing on [0, x2] and strictly decreasing on [x2, 1]. Since

f(0) = ln 2 + (2 p) ln 23

ln 2 + (2 r) ln 23

= 0

and f(1) = 0, we get f(x) min{f(0), f(1)} = 0.Equality occurs for x = y = z = 1. Additionally, for p = ln 9ln 8

ln 3ln 2 and x y z, equalityholds again when x = 0 and y = z = 3

2.

Proposition 3.16 ([8]). Ifx1, x2, . . . , xn (n 4) are non-negative numbers such thatx1 + x2 + + xn = n, then

1

n + 1 x2x3 xn +1

n + 1 x3x4 x1 + +1

n + 1 x1x2 xn1 1.

Proof. Let x1 x2 xn and en1 =

1 + 1n1

n1. By the AM-GM inequality, we have

x2 xn

x2 + + xnn 1

n1

x1 + x2 + + xnn 1

n1= en1.

Hencen + 1 x2x3 xn n + 1 en1 > 0,

and all denominators of the inequality are positive.

Case x1 = 0. It is easy to show that the inequality holds.

Case x1 > 0. Suppose that x1x2 xn = (n + 1)r = constant, r > 0. The inequality becomesx1

x1 r +x2

x2 r + +xn

xn r n + 1,



20/21

20 VASILE CRTOAJE

or1

x1 r +1

x2 r + +1

xn r 1

r.

By the AM-GM inequality, we have

(n + 1)r = x1x2 xn x1 + x2 + + xnn

n= 1,

hence r 1n+1

. From xn < x1 + x2 + + xn = n < n + 1 1r , we get xn < 1r . Therefore,we have r < xi < 1r for all numbers xi.

We will apply now Corollary 1.7 to the function f(u) = 1ur , u > r. We have f

(u) = 1(ur)2

and

g(x) = f

1

x

=

x2

(1 rx)2 , g(x) =

4rx + 2

(1 rx)4 .Since g(x) > 0, g(x) is strictly convex on

r, 1

r

. According to Corollary 1.7, if0 x1

x2

xn such that for x1 + x2 +

+ xn = constant and x1x2

xn = constant, then the

sum f(x1) + f(x2) + + f(xn) is minimal when x1 x2 = x3 = = xn. Thus, to provethe original inequality, it suffices to consider the case x1 = x and x2 = x3 = = xn = y,where 0 < x 1 y and x + (n 1)y = n. We leave ending the proof to the reader. Remark 3.17. The inequality is a particular case of the following more general statement:

Let n 3, en1 =

1 + 1n1

n1, kn =

(n1)en1nen1 and let a1, a2, . . . , an be non-negative

numbers such that a1 + a2 + + an = n.(a) Ifk kn, then

1

k a2a2 an +1

k a3a4 a1 + +1

k a1a2 an1 n

k 1;

(b) Ifen1 < k < kn, then1

k a2a3 an +1

k a3a4 a1 + +1

k a1a2 an1 n 1

k+

1

k en1 .

Finally, we mention that many other applications of the EV-Method are given in the book [2].

REFERENCES

[1] V. CRTOAJE, A generalization of Jensens inequality, Gazeta Matematica A, 2 (2005), 124138.

[2] V. CRTOAJE, Algebraic Inequalities - Old and New Methods, GIL Publishing House, Romania,2006.

[3] V. CRTOAJE, Two generalizations of Popovicius Inequality, Crux Math., 31(5) (2005), 313318.[4] V. CRTOAJE, Solution to problem 2724 by Walther Janous, Crux Math., 30(1) (2004), 4446.

[5] V. CRTOAJE, Problem 10528, A.M.M, 103(5) (1996), 428.

[6] V. CRTOAJE, Mathlinks site, http://www.mathlinks.ro/Forum/viewtopic.php?t=123604, Dec 10, 2006.

[7] V. CRTOAJE, Mathlinks site, http://www.mathlinks.ro/Forum/viewtopic.php?t=56732, Oct 19, 2005.

[8] V. CRTOAJE, Mathlinks site, http://www.mathlinks.ro/Forum/viewtopic.php?t=18627, Jan 29, 2005.

[9] V. CRTOAJE, Mathlinks site, http://www.mathlinks.ro/Forum/viewtopic.php?t=14906, Aug 04, 2004.

http://jipam.vu.edu.au/http://www.mathlinks.ro/Forum/viewtopic.php?t=14906http://www.mathlinks.ro/Forum/viewtopic.php?t=14906http://www.mathlinks.ro/Forum/viewtopic.php?t=18627http://www.mathlinks.ro/Forum/viewtopic.php?t=18627http://www.mathlinks.ro/Forum/viewtopic.php?t=56732http://www.mathlinks.ro/Forum/viewtopic.php?t=56732http://www.mathlinks.ro/Forum/viewtopic.php?t=123604http://www.mathlinks.ro/Forum/viewtopic.php?t=123604


21/21


[10] V. CRTOAJE, Mathlinks site, http://www.mathlinks.ro/Forum/viewtopic.php?t=19076, Nov 2, 2004.

[11] V. CRTOAJE, Mathlinks site, http://www.mathlinks.ro/Forum/viewtopic.php?t=21613, Feb 15, 2005.

[12] W. JANOUS AN D V. CRTOAJE, Two solutions to problem 2664, Crux Math., 29(5) (2003), 321323.

[13] D.S. MITRINOVIC, J.E. PECARIC AN D A.M. FINK, Classical and New Inequalities in Analysis,Kluwer, 1993.

[14] J. SURANYI, Miklos Schweitzer Competition-Hungary. http://www.mathlinks.ro/Forum/viewtopic.php?t=14008, Jul 11, 2004.

[15] http://www.mathlinks.ro/Forum/viewtopic.php?t=36346, May 10, 2005.

[16] http://www.mathlinks.ro/Forum/viewtopic.php?p=360537, Nov 2, 2005.
http://www.mathlinks.ro/Forum/viewtopic.php?p=360537http://www.mathlinks.ro/Forum/viewtopic.php?t=36346http://www.mathlinks.ro/Forum/viewtopic.php?t=14008http://www.mathlinks.ro/Forum/viewtopic.php?t=14008http://www.mathlinks.ro/Forum/viewtopic.php?t=21613http://www.mathlinks.ro/Forum/viewtopic.php?t=21613http://www.mathlinks.ro/Forum/viewtopic.php?t=19076http://www.mathlinks.ro/Forum/viewtopic.php?t=19076

Documents

The Equal Variable Method by Vasile Cirtoaje