Inequality with exponents $x^x+y^y \ge x^y +y^x$ Let $x,y$ be positive numbers. Prove that $x^x+y^y \ge x^y +y^x$. 

This question appeared in Summer 1991 Russian Olympiad team test. Apparently, I tried to come up with different approach such as Jensen, Karamata's inequality and nothing works so far. I just need a discussion here. Hints are not necessary.
without loss of generality we only prove by $0\le y\le x\le 1$,let
$$f(a)=a^{bx}-a^{by},x\ge a\ge y,1\ge bx-by\ge 0$$
then 
$$(a^{bx-by})'_{a}=\dfrac{bx-by}{a}\cdot a^{bx-by}&gt;0,\Longrightarrow a^{bx-by}\ge y^{bx-by}\cdots (1)$$
since use $AM-GM$ inequality,we have
$$y^{1+y-x}1^{x+xy-y^2-y}\le\left(\dfrac{x}{1+xy-y^2}\right)^{1+xy-y^2}\le x\cdots (2)$$
and
$$f'(a)=\dfrac{bx}{a}\cdot a^{bx}-\dfrac{by}{a}\cdot a^{by}=\dfrac{ba^{by}}{a}(xa^{b(x-y)}-y)=$$
so use $(1),(2)$ we have
$$f'(a)\ge ba^{by-1}(xy^{b(x-y)}-y)=ba^{by-1}y^{b(x-y)}(x-y^{1-b(x-y)})\ge 0
$$
so
$$f(x)\ge f(y)\Longrightarrow x^{bx}+y^{by}\ge x^{by}+y^{bx}$$
let $b=1$

we have
$$x^x+y^y\ge x^y+y^x$$
Use induction, taking the base case x=0