Inequality with exponents $x^x+y^y \ge x^y +y^x$

Question

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. I tried to come up with different approach such as Jensen or Karamata's inequality, but nothing worked so far. I just need a discussion here. Hints are not necessary.

Answer 1

Use induction, taking the base case x=0

Answer 2

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.

Answer 3

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 $$x^x+y^y\ge x^y+y^x$$