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This question already has an answer here:

Prove that: $$e ^ π > π ^ e.$$ Hint: Take the natural log of both sides and try to define a suitable function that has the essential properties that yields the above inequality

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marked as duplicate by user127.0.0.1, 6005, user99914, Stefan Hansen, Claude Leibovici Feb 26 '14 at 8:13

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Put $f(x) = \frac{ \ln x }{x} $. then $f' = \frac{1 - \ln x}{x^2} $. $f' = 0 \iff 1 - \ln x = 0 \iff x = e $. Hence $\sup_x f(x) = f(e) = \frac{ \ln e }{e} $. In particular, this must be $> \frac{ \ln \pi}{ \pi} $.

$$ \therefore \frac{\ln e}{e} > \frac{ \ln \pi}{\pi} \iff e^\pi > \pi^e$$

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Another Hint: $$ x^y > y^x \iff y \log x > x \log y \iff \frac{\log y}{y} < \frac{\log x}{x}. $$

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  • $\begingroup$ Sorry but one can't see, only by looking at it without graph, that $\frac {ln(e)}{e}>\frac{ln(\pi)}{\pi}$ the difference is really small $\endgroup$ – Bman72 Feb 26 '14 at 7:46
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    $\begingroup$ @Ale You don't just look at it, you need to use the first hint you had. Which is, find a function. My hint is intended to make it clear that the desired function is $f(x) = \frac{\log x}{x}$. $\endgroup$ – 6005 Feb 26 '14 at 7:53

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