Let $x>y>0$. Prove that $$x^{y^x}>y^{x^y}.$$ My attempts:

- Let $1>x>y>0$.

In this case it's enough to prove that $$y^x<x^y$$ or $$x\ln\frac{1}{y}>y\ln\frac{1}{x},$$ which is obvious;

- $x\geq1>y>0$.

In this case our inequality is obviously true;

- $e\geq x>y>1.$.

We need to prove that $f(x)\geq0,$ where $$f(x)=x\ln{y}-y\ln{x}+\ln\ln{x}-\ln\ln{y}.$$ Now, $$f'(x)=\ln{y}-\frac{y}{x}+\frac{1}{x\ln{x}}.$$ Let $h(y)=\ln{y}-\frac{y}{x}+\frac{1}{x\ln{x}}.$

Thus, $$h'(y)=\frac{1}{y}-\frac{1}{x}>0,$$ which says $$h(y)>h(1)=-\frac{1}{x}+\frac{1}{x\ln{x}}=\frac{1-\ln{x}}{x\ln{x}}\geq0.$$ Id est, $f$ increases and $$f(x)>f(y)=0;$$ 4. $x>y\geq e$.

Since $$\left(\frac{\ln{x}}{x}\right)'=\frac{1-\ln{x}}{x^2}\leq0$$ for all $x\geq e,$ we obtain: $$f(x)=xy\left(\frac{\ln{y}}{y}-\frac{\ln{x}}{x}\right)+\ln\ln{x}-\ln\ln{y}>0;$$ 5. $x\geq e>y>1.$

In this case I am stuck.

Thank you!