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. 
 A: 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 
$$(a^{bx-by})'_{a}=\dfrac{bx-by}{a}\cdot a^{bx-by}>0,\Longrightarrow a^{bx-by}\ge y^{bx-by}\cdots (1)$$
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
$$x^x+y^y\ge x^y+y^x$$
A: Let $x\geq y$.
We'll consider two cases.


*

*$x\geq1$.


Hence, $$x^x+y^y=x^y\left(x^{x-y}-1\right)+x^y+y^y\geq y^y\left(x^{x-y}-1\right)+x^y+y^y\geq$$
$$\geq y^y\left(y^{x-y}-1\right)+x^y+y^y\geq x^y+y^x.$$
2. $1>x\geq y>0.$
Let $f(t)=t^x-t^y$, where $t\in[y,1).$
Thus, by Bernoulli we obtain:
$$f'(t)=xt^{x-1}-yt^{y-1}=t^{y-1}\left(xt^{x-y}-y\right)\geq$$
$$\geq t^{y-1}\left(xy^{x-y}-y\right)=t^{y-1}y^{x-y}\left(x-y^{1-x+y}\right)=$$
$$=t^{y-1}y^{x-y}\left(x-(1+(y-1))^{1-x+y}\right)\geq$$
$$\geq t^{y-1}y^{x-y}\left(x-(1+(y-1)(1-x+y))\right)=t^{y-1}y^{x-y}y(x-y)\geq0,$$
which gives $$f(x)\geq f(y)$$ or
$$x^x-x^y\geq y^x-y^y$$ or
$$x^x+y^y\geq x^y+y^x$$ and we are done!
A: Assume $x\ge y$ by symmetry. We want
$$x^x-x^y\ge y^x-y^y.$$
i.e. $$x^y(x^{x-y}-1)\ge y^y(y^{x-y}-1),$$
which is then easy to show by listing all possible cases according to how $x,y$ compare to 1. 
A: Use induction, taking the base case x=0
