Is there a good reason why $a^{2b} + b^{2a} \le 1$ when $a+b=1$? The following problem is not from me, yet I find it a big challenge to give a nice (in contrast to 'heavy computation') proof. The motivation for me to post it lies in its concise content.
If $a$ and $b$ are nonnegative real numbers such that $a+b=1$, show that  $a^{2b} + b^{2a}\le 1$.

Maybe there is a small trick yet.
For $a + b = 1$ we can write the sum as 
$$a^{2(1-a)} + b^{2(1-b)} = (\frac{a}{a^a})^2+ (\frac{b}{b^b})^2 $$.
Obviously the sum is 1 for $(a, b) = (0,1), (\frac{1}{2},\frac{1}{2}), (1, 0)$. The question is whether at $(\frac{1}{2},\frac{1}{2})$ there is a maximum or a minimum, i.e. whether the second derivative of
$$(\frac{x}{x^x})^2+ (\frac{1-x}{(1-x)^{1-x}})^2$$
is positive or negative. (By symmetry only minimum or maximum can occur there.)
Using the fact that, for $x = \frac{1}{2}$,
$$\frac{d^2}{dx^2}(\frac{1-x}{(1-x)^{1-x}}) =  \frac{d^2}{(-dx)^2}(\frac{x}{x^x})^2 =  \frac{d^2}{dx^2}(\frac{x}{x^x})^2$$
it is sufficient to prove
$$2\frac{d^2}{dx^2}(\frac{x}{x^x})^2 < 0$$
for $x = \frac{1}{2}$ which in fact is easily demonstrated.
