Jtdylktuy

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 $0le yle xle 1$,let
$$f(a)=a^{bx}-a^{by},xge age y,1ge bx-byge 0$$
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
$$y^{1+y-x}1^{x+xy-y^2-y}leleft(dfrac{x}{1+xy-y^2}right)^{1+xy-y^2}le xcdots (2)$$
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
$$x^x+y^yge x^y+y^x$$

Let $xgeq y$.

We'll consider two cases.

1. 
   $xgeq1$.

Hence, $$x^x+y^y=x^yleft(x^{x-y}-1right)+x^y+y^ygeq y^yleft(x^{x-y}-1right)+x^y+y^ygeq$$
$$geq y^yleft(y^{x-y}-1right)+x^y+y^ygeq x^y+y^x.$$
2. $1>xgeq y>0.$

Let $f(t)=t^x-t^y$, where $tin[y,1).$

Thus, by Bernoulli we obtain:
$$f'(t)=xt^{x-1}-yt^{y-1}=t^{y-1}left(xt^{x-y}-yright)geq$$
$$geq t^{y-1}left(xy^{x-y}-yright)=t^{y-1}y^{x-y}left(x-y^{1-x+y}right)=$$
$$=t^{y-1}y^{x-y}left(x-(1+(y-1))^{1-x+y}right)geq$$
$$geq t^{y-1}y^{x-y}left(x-(1+(y-1)(1-x+y))right)=t^{y-1}y^{x-y}y(x-y)geq0,$$
which gives $$f(x)geq f(y)$$ or
$$x^x-x^ygeq y^x-y^y$$ or
$$x^x+y^ygeq x^y+y^x$$ and we are done!

Assume $xge y$ by symmetry. We want
$$x^x-x^yge y^x-y^y.$$
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.

Use induction, taking the base case x=0