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 $f(e) > f(\pi)$.
$e^{\frac{1}{e}} > \pi^{\frac{1}{\pi}}$.
$e^{\pi} > \pi^{e}$.

$</code> ... <code>$</code>...<code>."> Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in $</span> ... <span>$ or $</span> ... <span>$ to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestA 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 $e^\pi>\pi^e$.

Log in to reply

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

Log in to reply

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.

Log in to reply

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

Log in to reply

Log in to reply

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.Log in to reply

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

Log in to reply

Brilliant! Great solution.

Log in to reply