Question

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

Answer 1

We have, $1=a+b=a^{a+b}{b}^{a+b}=a^a{b}^b+b^a{b}^b$
$\therefore 1-a^a{b}^b-a^b{b}^a=a^a{b}^b+b^a{b}^b-a^b{b}^a=(a^a-b^a)(a^b-b^b)$
Now if a $\le$ b, then $a^a\le b^a$ and $a^b\le b^b$. If a $\ge$ b, then $a^a\ge b^a$ and $a^b\ge b^b$. Hence the product is non negative for all positive a and b.
$\therefore a^a{b}^b+a^b{b}^a\le 1$