$e^\pi - \pi^e < 1$?

Question

We have Comparing $\pi^e$ and $e^\pi$ without calculating them but it doesn't give an approximation of the actual difference. Is there a way without calcualting an approximation of them to prove $e^\pi - \pi^e < 1$ ?

Answer 1

If that can help:

Let $f(x):=e^x-x^e$. This function has a minimum at $x=e$ (double root), and the second order Taylor development is

$$y\approx g(x):=e^{e-1}(x-e)^2.$$

This approximation exceeds $f$, but we still have $g(\pi)<1$.

In blue, $f$, in black, $g$.