Prove that if $e<y<x$ then $x^{y}<y^{x}$

Question

Prove that if $e<y<x$ then $x^{y}<y^{x}$

My try:
I tried to use Taylor's theorem: $$y^{x}-x^{y}=e^{xlny}-e^{ylnx}=1+xlny+o(xlny)-1-ylnx-o(ylnx)=$$ $$=lny^{x}-lnx^{y}+o(xlny)-o(ylnx)=ln\frac{y^{x}}{x^{y}}+o(xlny)-o(ylnx)$$Hovewer I have a problem to show that $y^{x}-x^{y}>0$ because I can't say that $ln\frac{y^{x}}{x^{y}}>0$ because then I use with theses.

Have you some idea?

Answer 1

$e^{x\ln(y)}<e^{y\ln(x)}$ is equivalent to ${{\ln(x)}\over x}<{{\ln(y)}\over y}$

Let $f(x)={{\ln(x)}\over x}$, $f'(x)={{1-\ln(x)}\over x^2}<0$ if $x>e$, so if $e<y<x$, $f(x)<f(y)$

Answer 2

The hint:

Prove that $f(x)=\frac{x}{\ln{x}}$ increases for $x>e$.