What conditions of $x$ such that $e^{x^e} \ge x^{e^x}$?

Question

What conditions of $x$ such that $e^{x^e} \ge x^{e^x}$ ?

Answer 1

One might start by noticing that $x^{e^x}$ is undefined for $x<0$. Assuming we are looking at the domain $x\ge0$, one can see that for $x>e$,

$$x^{e^x}>e^{e^x}>e^{x^e}$$

So if it were to hold true, $x\le e$. At $x=e$, they are trivially equal.

One might also notice that

$$x^{e^x}=e^{e^x\ln(x)}$$

So the problem reduces to solving

$$e^x\ln(x)\le x^e$$

When $x\le1$, we trivially have

$$e^x\ln(x)\le0\le x^e$$

For $1<x<e$, take the log of both sides to get

$$x+\ln(\ln(x))\le e\ln(x)$$

This may be done by considering

$$f(x)=e\ln(x)-x-\ln(\ln(x))$$

And noting that

$$f(e)=0,f'(x)<0\forall x\in(1,e)$$

Answer 2

Define for $x>0$:

$$f(x)=\ln\frac{e^{x^e}} {x^{e^x}}=x^e-e^x\ln x$$

A plotting of this function suggests that $f(x)\ge e$ for some interval $(0,a]$ where $a\approx 0.76$.