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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?

Jeel Shah
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N3buchadnezzar
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    What a nice inequality cascade! – Hypergeometricx Nov 20 '14 at 15:04
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    You can see that left and right inequalities are obvious for the one in the middle just consider $x^{\frac{1}{x}}$. Differentiate it. – happymath Nov 20 '14 at 15:05
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    The only difficult part here is proving $\sqrt[\pi]{\pi}<\sqrt[e]{e}$, which is equivalent to proving ${\pi}^{e}<{e}^{\pi}$. The answer to that can be found at http://math.stackexchange.com/a/9459/131263. – barak manos Nov 20 '14 at 15:14

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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.

Przemysław Scherwentke
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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 $\pi^e<e^{\pi}$.

Glorfindel
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Rory Daulton
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    I really liked the idea from the other thread. We know that $\pi > e$ and $e^x > 1+x$ for $x>0$. From the first bit we know that $\pi/e>1$ and $e^{\pi/e-1}>\pi/e$. Which implies that $e^{\pi/e}>\pi$. Raising both sides to $e$ completes the proof. – N3buchadnezzar Nov 20 '14 at 15:54
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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 $x=e$.

Akiva Weinberger
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  • The inequality I gave at the start, $e^x\ge x+1$, actually works pretty well as a definition of $e$. (It allows you to prove many $e$-related results without calculus.) As I've shown, this is equivalent to saying that the maximum of $\sqrt[x]x$ is at $x=e$. That means that that also works pretty well as a definition of $e$. – Akiva Weinberger Apr 26 '15 at 19:08