TITLE: $\forall x,y>0, x^x+y^y \geq x^y + y^x$
QUESTION [16 upvotes]: Prove that $\forall x,y>0, x^x+y^y \geq x^y + y^x$

A friend of mine told me none of the teachers in my school have succeeded in proving this seemingly simple inequality (it was asked at an oral exam last year). I tried it myself, but I made no significant progress.

REPLY [10 votes]: WLOG, let $x \ge y$. 
Case 1: Let $x \ge 1$, then for $t > 0$, $x^t \log x \ge y^t \log y$
$$\implies \int_y^x \left(x^t \log x - y^t \log y \right)dt \ge 0 \implies (x^x - x^y) - (y^x-y^y) \ge  0$$
Added
Case 2: For $x< 1$ - let $r = \dfrac{x}y\ge 1$. Then the inequality is:
$$\frac{y^x}{x^y+y^x}r^x+\frac{x^y}{x^y+y^x}\frac1{r^y} \ge 1$$
Using weighted AM-GM, this reduces to showing: 
$$xy^x \ge yx^y \iff \frac{\log x}{1-x} \ge \frac{\log y}{1-y}$$
which is obvious.