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- Feb 14, 2012

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Let $a$ and $b$ be positive real numbers such that $a+b=1$. Prove that $a^ab^b+a^bb^a\le 1$.

- Thread starter anemone
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- Thread starter
- Admin
- #1

- Feb 14, 2012

- 3,802

Let $a$ and $b$ be positive real numbers such that $a+b=1$. Prove that $a^ab^b+a^bb^a\le 1$.

- Mar 31, 2013

- 1,331

$1= a+ b = a^{a+b} + b^{a+b}$

So $1- (a^ab^b + a^b b^a)$

$= a^{a+b} + b^{a+b} - (a^ab^b + a^b b^a)$

$= a^a(a^b-b^b) + b^a(b^b-a^b) = (a^a - b^a)(a^b - b^b)$

For a > b both the terms are non -ve so we have and if b > a then both terms are -ve and hence above is positive

$1- (a^ab^b + a^b b^a) >=0$ and hence the result