How to prove $\sqrt[\pi]{e} < \sqrt[\pi]{\pi}<\sqrt[e]{e}< \sqrt[e]{\pi}$ I was given a challenge of sorting the following numbers. $\Large\sqrt[\pi]{e} < \sqrt[\pi]{\pi}<\sqrt[e]{e}< \sqrt[e]{\pi}$. After some work I was able to figure out the order. How can one prove this by hand?
 A: If you take each expression to the power $e \cdot \pi$, most of the order is obvious. The only non-obvious part is comparing $e^{\pi}$ and $\pi^e$ (without a calculator).
That last is a classic problem with many solutions. There are examples here and here. Here is another way, given in "Calculus: Graphical, Numerical, Algebraic" by Finney et al.
Consider the line through the origin tangent to the graph of $y=\ln x$, which has slope $1/e$. Find the equation of the tangent line and use the concavity downward of $\ln x$ to see the tangent line is above the graph. This shows that $\ln(x^e)<x$ for all positive $x \ne e$. Therefore, $x^e<e^x$ for all positive $x \ne e$, so $\pi^e<e^{\pi}$.
A: Clearly, $\sqrt[\pi]e<\sqrt[\pi]\pi$, and $\sqrt[e]e<\sqrt[e]\pi$. (That's because $e<\pi$.)
The only hard part is showing that $\sqrt[\pi]\pi<\sqrt[e]e$.
I'm going to do this by using my favorite inequality involving $e$:
$$e^x\ge x+1$$
with equality iff $x=0$. (Look at a graph to see why this is. It's basically another way of saying that the derivative of $e^x$ at $0$ is $1$.)
Substitute in $x=\frac\pi e-1$. You'll see why I'm using this particular value in a moment.
\begin{align}
e^{\pi/e-1}&>\left(\frac\pi e-1\right)+1\\
e^{\pi/e-1}&>\frac\pi e\\
e^{\pi/e}&>\pi\\
e^{1/e}&>\pi^{1/\pi}\\
\sqrt[e]e&>\sqrt[\pi]{\pi}\tag*{$\blacksquare$}
\end{align}
You might have noticed that, in the above argument, there's nothing special about $\pi$. In fact, the above argument can be modified to show that $\sqrt[x]x\le\sqrt[e]e$ for any $x$ (with equality iff $x=e$). In other words, the maximum of $\sqrt[x]x$ is at $x=e$.
A: HINT: The only non-obvious inequality is $ \sqrt[\pi]{\pi}<\sqrt[e]{e}$. Taking logarithm we obtain
$$
\frac1\pi\ln\pi<\frac1e\ln e
$$
and it remains to show, that 
$$
\frac{\ln x}{x}
$$
has its maximum at $x=e$, but it is a standard calculation.
