Prove that $e ^ π$ > $π ^ e$.

Question

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

Answer 1

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} $.

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

Answer 2

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