if $a^b=b^a, 0<b<a$ then $ab$ and $e^2$ which is bigger?
I try $$b^a=a^b\Longrightarrow a\ln{b}=b\log{a}\Longrightarrow \dfrac{\ln{a}}{a}=\dfrac{\ln{b}}{b}$$,where $\ln{x}=\log{x}=\log_{e}{x}$
consider the function $$f(x)=\dfrac{\ln{x}}{x},x>0,\Longrightarrow f'(x)=\dfrac{1-\ln{x}}{x^2},x>0$$if $f'(x)=0$,then $x=e$.
The solutions $0<b<a$ to the identity $a^b=b^a$ can be parametrized by $x>0$ as $$b:=(1+1/x)^x\qquad a:=(1+1/x)^{x+1}$$ hence $$\sqrt{ab}=(1+1/x)^{x+1/2}$$ The "Note" to the question Find the smallest $a>0$ such that $(1+\frac{1}{x})^{x+a}>e$ for all $x\geq 1\,$. (valid for $x>0$) shows that $$(1+1/x)^{x+1/2}>e$$ Thus, $$\sqrt{ab}>e$$
