A question comparing $\pi^e$ to $e^\pi$ [duplicate]

Solution 1:

Consider the function $x^{\frac{1}{x}}$. Differentiating gives $x^{\frac{1}{x}}(\frac{1}{x^2})(1-\ln x)$, so the function attains its global maximum at $x=e$.

Thus $e^{\frac{1}{e}} \geq \pi^{\frac{1}{\pi}}$, and it is clear that the inequality is strict, so $e^{\pi}>\pi^{e}$.

Solution 2:

If you know Taylor expansion: then $$e^x=1+x+\frac{x^2}{2!}+...$$ We can get (Or you may take derivative to prove it) $$e^{x}>1+x, \forall x>0$$ Then set $$x=\frac{\pi}{e}-1>0$$ We get $$e^{\frac{\pi}{e}-1}>1+\frac{\pi}{e}-1\Leftrightarrow \frac{e^{\frac{\pi}{e}}}{e}>\frac{\pi}{e}\Leftrightarrow e^{\frac{\pi}{e}}>\pi\Leftrightarrow e^{\pi}>\pi^e$$

Solution 3:

Note that if $f(x) = \dfrac{\log(x)}x$, we have $$f'(x) = \dfrac{x \times \dfrac1x - \log(x)}{x^2} \begin{cases} < 0 & \forall x > e\\ > 0& \forall x < e\end{cases} $$ Hence, $f(x)$ is a strictly montone decreasing function for $x \geq e$ and strictly montone increasing function for $x \leq e$. Since $\pi > 3 > e$, we have $$f(\pi) < f(e) \implies \dfrac{\log(\pi)}{\pi} < \dfrac{\log(e)}{e} \implies \pi^e < e^{\pi} \implies \pi^{1/\pi} < e^{1/e}$$