eπ>πe e^ \pi > \pi ^ e .

Prove that eπ>πe 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, 2 months ago

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

It derivative is : f(x)=1logxx2f'(x)= \frac{1-\log x}{x^2}, therefore f(e)f(e) is the maximum.

Then : f(e)>f(π)f(e)>f(\pi), which means that : πloge>elogπeπ>πe\pi \log e >e\log \pi \Leftrightarrow e^{\pi} >\pi^e.

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

Haroun Meghaichi - 5 years, 2 months ago

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Brilliant! Great solution.

Carlos E. C. do Nascimento - 5 years, 2 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 - 5 years, 1 month ago

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I took the function f(x)=x1xf(x) = x^{\frac{1}{x}} and found that it has a maximum at x=ex = e.So f(e)>f(π)f(e) > f(\pi). e1e>π1π e^{\frac{1}{e}} > \pi^{\frac{1}{\pi}} . eπ>πee^{\pi} > \pi^{e}.

Eddie The Head - 5 years, 2 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 - 5 years, 2 months ago

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

Eddie The Head - 5 years, 2 months ago

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@Eddie The Head Hey, me too.

Michael Mendrin - 5 years, 2 months ago

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A well-known and important inequality: en>1+ne^n>1+n given n>0n>0. Now if we let n=πe1n=\dfrac \pi e-1 we will immediately get after rearranging our inequality eπ>πee^\pi>\pi^e.

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