Is $e^{\pi}$ greater than $\pi^{e}$?

[Ackbach]

Here is this week's POTW:

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Prove that $e^{\pi}>\pi^{e}$.

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[Ackbach]

Congratulations to Rido12 for his correct solution! It is posted below.

Spoiler

To prove $e^{\pi}>\pi^e$, let us first generalize it: $e^x>x^e \iff x>e \ln x \iff \frac{x}{\ln x}-e>0$.

Consider the function $f(x)=\frac{x}{\ln x}-e$ with derivative $f'(x)=\frac{\ln x -1}{\ln^2 x}$. The critical values of $f$ are at $x=1$ and $x=e$, and $f'(x)<0$ for $x \in (0,e)$ and $f'(x)>0$ for $x>e$. The function obtains a minimum at $x=e$ and hence is monotonically increasing for $x>e$.
Since $f(x)>f(e)$ for $x>e$ and $\pi>e$, then $f(\pi)>f(e)=0$.
$$ f(\pi)>0 \iff \frac{\pi}{\ln \pi}>e\iff e^{\pi}>\pi^e$$ $$, as required.