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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4$f(x)=x^{-1}\log x$, optimize you must. – Pedro Feb 26 '14 at 07:12
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1One of those questions that gets asked many, many times: see, for instance http://math.stackexchange.com/questions/7892/comparing-pie-and-e-pi @PedroTamaroff Optimization at the foot of Master Yoda learn you did? – colormegone Feb 26 '14 at 07:19
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http://math.stackexchange.com/questions/337565/a-question-comparing-pie-to-e-pi – lab bhattacharjee Feb 26 '14 at 07:23
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duplicate questions on the prowl – Apurv Feb 26 '14 at 07:31
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I love this link: https://arxiv.org/pdf/1806.03163.pdf – insipidintegrator Jul 01 '22 at 14:52
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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: $$ x^y > y^x \iff y \log x > x \log y \iff \frac{\log y}{y} < \frac{\log x}{x}. $$
Caleb Stanford
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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 – Bman72 Feb 26 '14 at 07:46
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2@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 $f(x) = \frac{\log x}{x}$. – Caleb Stanford Feb 26 '14 at 07:53