Prove that \( e^ \pi > \pi ^ e \).

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## Comments

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TopNewestTake 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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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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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 \(e^\pi>\pi^e\).

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