# Prove that $e ^ π$ > $π ^ e$. [duplicate]

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

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

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 $> \frac{ \ln \pi}{ \pi}$.
$$\therefore \frac{\ln e}{e} > \frac{ \ln \pi}{\pi} \iff e^\pi > \pi^e$$
Another Hint: $$x^y > y^x \iff y \log x > x \log y \iff \frac{\log y}{y} < \frac{\log x}{x}.$$
• 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 7:46
• @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 $f(x) = \frac{\log x}{x}$. – 6005 Feb 26 '14 at 7:53