Last week, Worldwide Center of Mathematics posted this problem on their Facebook page. So, I felt challenged. The solution turned to be quite simple, but that's not the point. The point is, this inequality simply looks beautiful.

Long story short, the task is to prove the following $$e^{\pi} > \pi^{e}$$

I will use few tricks, in form of this well known inequality $$x-1 \geq \ln(x), \forall x >0$$ and the fact that $x=\ln(\pi) > \ln(e)=1$. In fact this inequality is strict for $\forall x > 1$

As a result $$\ln(\pi)-1 > \ln(\ln(\pi)) \Leftrightarrow \ln(\pi) > 1+\ln(\ln(\pi)) $$

Luckily, exponential function $f(x)=e^x$ is strictly ascending, thus: $$e^{\ln(\pi)} > e^{1+\ln(\ln(\pi))} \Leftrightarrow \pi > e \cdot \ln(\pi) \Leftrightarrow \pi > \ln(\pi^e)$$

Using the exponential function trick again $$e^{\pi} > e^{\ln(\pi^e)} \Leftrightarrow e^{\pi} > \pi^e$$