From: leonsotelo To: leonsotelo

Date: 1/22/2005 1:18:39 AMSubjec t : a+b=1

173.Suppose that a and b are positive real numbers for which a + b = 1. Prove that

a +1

a

2

+

b +1

b

2

25

2 .

Determine when equality holds.

Remark. Before starting, we note that when a + b = 1, a, b > 0, then ab 1/4. This is an immediate consequence of the arithmetic-geometric means inequality.

Solution 1. By the root-mean-square, arithmetic mean inequality, we have that

1

2

a +1

a

2

+

b +1

b

2

1

4

a +1

a

+

b +1

b

2

=1

4

1 +1

a+

1

b

2

=

1

4

1 +1

ab

2

1

452

as desired.

Solution 2. By the RMS-AM inequality and the harmonic-arithmetic means inequality, we have that

a2 + b2 + (1/a)2 + (1/b)2 1

2(a + b)2 +

1

2

1

a+

1

b

2

=1

2+ 2

(1/a) + (1/b)

2

2

1

2+ 2

4

(a+b)2=

17

2 ,

from which the result follows.

Solution 3.

a +1

a

2

+

b +1

b

2

= a2 + b2 +a2 + b2

a2b2+ 4

= (a + b)2 - 2ab +(a+b)2 - 2ab

a2b2+ 4

= 5 - 2ab +1

(ab)2-

2

ab

= 4 - 2ab +

1

ab- 1

2

4 - 2

1

4

+ (4 - 1)3 =25

2 .

Solution 4. [F. Feng] From the Cauchy-Schwarz and arithmetic-geometric means inequalities, we find that

a +1

a

2

+

b +1

b

2

[12 + 12]

=

a +1

a

2

+

b +1

b

2

[(a + b)2 + (a + b)2 ]

a +1

a

(a + b) +

b +1

b

(a + b)

2

=

(a + b)2 + 2 +

a

b+

b

a

2

[1 + 2 + 2]2 = 25 .

The desired result follows.

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