<s>[INST] if $x^y=y^x$ show that $x+y>2e$ 
Let $0<x<y$, such that
  $$x^y=y^x$$
  show that
  $$x+y>2e$$

Since $$y\ln{x}=x\ln{y}\Longrightarrow \dfrac{\ln{y}}{y}=\dfrac{\ln{x}}{x}$$
Let 
$$f(x)=\dfrac{\ln{x}}{x}\Longrightarrow f'(x)=\dfrac{1-\ln{x}}{x^2}$$
If $0<x<e$ then  $f'(x)>0$, if $x>e$,then $f'(x)<0$
$\Longleftrightarrow y-e+x-e>0$,maybe consider
$f(x-e)$ and $f(y-e)$,wihch bigger?
I just do it now 
Thanks in advance!
 [/INST] As  MathematicianByMistake answer,we want prove
$$g(r)=r^{\frac{1}{r-1}}+r^{\frac{r}{r-1}}=(x+1)^{\frac{1}{x}}+(x+1)^{\frac{x+1}{x}}=(x+1)^{\frac{1}{x}}(x+2),x>0$$
since
$$g'(x)=(x+1)^{\frac{1}{x}-1}\cdot\dfrac{x(x^2+2x+2)-(x^2+3x+2)\ln{(x+1)}}{x^2}$$
we only prove
$$h(x)=x(x^2+2x+2)-(x^2+3x+2)\ln{(x+1)}>0,x>0$$
since
$$h'(x)=3x(x+1)-(3x+2)\ln{(x+1)}>3x(x+1)-x(3x+2)=x>0$$
</s>