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

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This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

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## Comments

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

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

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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.

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

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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 ;)

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Brilliant! Great solution.

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