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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asked Feb 26, 2014 at 7:11
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4$\begingroup$ $f(x)=x^{-1}\log x$, optimize you must. $\endgroup$– Pedro ♦Feb 26, 2014 at 7:12
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$\begingroup$ One of those questions that gets asked many, many times: see, for instance math.stackexchange.com/questions/7892/comparing-pie-and-e-pi @PedroTamaroff Optimization at the foot of Master Yoda learn you did? $\endgroup$– colormegoneFeb 26, 2014 at 7:19
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$\begingroup$ math.stackexchange.com/questions/337565/… $\endgroup$– lab bhattacharjeeFeb 26, 2014 at 7:23
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$\begingroup$ duplicate questions on the prowl $\endgroup$– ApurvFeb 26, 2014 at 7:31
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$\begingroup$ I love this link: arxiv.org/pdf/1806.03163.pdf $\endgroup$– insipidintegratorJul 1, 2022 at 14:52
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2 Answers
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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: $$ x^y > y^x \iff y \log x > x \log y \iff \frac{\log y}{y} < \frac{\log x}{x}. $$
answered Feb 26, 2014 at 7:19
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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 $\endgroup$– Bman72Feb 26, 2014 at 7:46
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2$\begingroup$ @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 $f(x) = \frac{\log x}{x}$. $\endgroup$ Feb 26, 2014 at 7:53