# $e^ \pi > \pi ^ e$.

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

This is a list of Calculus proof based problems that I like. Please avoid posting complete solutions, so that others can work on it. Note by Calvin Lin
5 years, 4 months ago

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A well-known and important inequality: $e^n>1+n$ given $n>0$. Now if we let $n=\dfrac \pi e-1$ we will immediately get after rearranging our inequality $e^\pi>\pi^e$.

I took the function $f(x) = x^{\frac{1}{x}}$ and found that it has a maximum at $x = e$.So $f(e) > f(\pi)$. $e^{\frac{1}{e}} > \pi^{\frac{1}{\pi}}$. $e^{\pi} > \pi^{e}$.

- 5 years, 3 months ago

This is the way I did it a long time ago. This one is an oldie. But this approach takes longer than Haroon's proof.

- 5 years, 3 months ago

Yeah..I also solved this one a long time ago ....before i joined Brilliant!!

- 5 years, 3 months ago

Hey, me too.

- 5 years, 3 months ago

Take the function $f:\mathbb{R_+}\mapsto \mathbb{R}$, defined by : $f(x)=\frac{\log x}{x}$.

It derivative is : $f'(x)= \frac{1-\log x}{x^2}$, therefore $f(e)$ is the maximum.

Then : $f(e)>f(\pi)$, which means that : $\pi \log e >e\log \pi \Leftrightarrow e^{\pi} >\pi^e$.

How did I choose this function : just try to get $\pi$ is one side and $e$ is another to get the idea.

- 5 years, 4 months ago

This solution exists in the math book of highschool's terminal year in Algeria. I loved it ;)

- 5 years, 2 months ago

Brilliant! Great solution.

- 5 years, 4 months ago