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

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
3 years, 8 months ago

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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.

Haroun Meghaichi - 3 years, 8 months ago

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

Zakaria Sellami - 3 years, 7 months ago

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Brilliant! Great solution.

Carlos E. C. do Nascimento - 3 years, 8 months ago

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

Eddie The Head - 3 years, 8 months ago

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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.

Michael Mendrin - 3 years, 8 months ago

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

Eddie The Head - 3 years, 8 months ago

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@Eddie The Head Hey, me too.

Michael Mendrin - 3 years, 8 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 \(e^\pi>\pi^e\).

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