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The other day, a math teacher at my college gave me a challenge problem: Prove that $$\pi^e < e^{\pi}$$ without using a calculator. The next day, I found a valid proof, but I used a log table instead of a calculator, so my proof is hardly satisfying. I went looking for another proof and I came up with something that might not be legitimate. Here's my proof:

Suppose that is an $x$ such that $x^e < e^x$

Then taking the natural log of both sides,

$$\ln(x^e) < \ln(e^x)$$

$$e\ln(x) < x$$

$$\ln(x) < \frac{x}{e}$$

Differentiate both sides,

$$\frac{1}{x} < \frac{1}{e}$$


Thus, $x^e < x^e$, for $x > e$

Since $\pi > e$, therefore $\pi^e < e^{\pi}$

I think my proof is good except perhaps where I differentiated both sides of the inequality. I know that there are many cases where this is invalid, but I'm not sure about this case.


marked as duplicate by Mark Viola, user147263, user99914, Mike Pierce, Math1000 Jul 5 '15 at 2:18

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  • $\begingroup$ you can modify your proof by using the function $f(x)=e^x-x^e$. Now feel free to differentiate... $\endgroup$ – user72012 Jul 4 '15 at 18:00
  • $\begingroup$ Notice in your relation $1/x< 1/e$ what happens for, e.g., $x=0.1$. For the relation to be preserved, you need the derivative in the RH side to be always at least as large as the one on the LH side. $\endgroup$ – Gary. Jul 4 '15 at 18:02
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    $\begingroup$ See this for other approaches. $\endgroup$ – David Mitra Jul 4 '15 at 18:03
  • $\begingroup$ This question has already many answers at MSE, see here. $\endgroup$ – Dietrich Burde Jul 4 '15 at 18:14

Your differentiating is not valid. Comparison of gradients is not enough for comparison of functions. I'll provide two alternatives to prove it below:

Taylor Expansion:

Using the Taylor expansion of $e^x$, $$e^x=1+x+\frac{x^2}{2!}+ \cdots$$

so that $e^x > 1 + x$, as long $x \neq 0$. Then letting $x = \frac{\pi}{e} - 1$ we have $$e^{\pi/e -1} > \pi/e,$$

and so, multiplying both sides by $e$ give us

$$e^{\pi/e} > \pi.$$


$$e^{\pi} > \pi^e.$$

Calculus-based alternative:

Consider the function $x^{\frac{1}{x}}$. Differentiating this function yields $$x^{\frac{1}{x}}\left(\frac{1}{x^2}\right)(1-\ln x)$$ We can see that this function attains its global maximum at $x=e$ by setting the derivative to $0$ and solving.

Hence we have $$e^{\frac{1}{e}} > \pi^{\frac{1}{\pi}}$$ so we get $$e^{\pi}>\pi^{e}.$$

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    $\begingroup$ That is a cute and nice method. +1 $\endgroup$ – Landon Carter Jul 4 '15 at 18:00

You cannot differentiate like that. That's very wrong.

Hint: consider the function $f(x)=x^{1/x},\space x>0$. Show that $f$ attains its maximum when $x=e$. And you are done.

  • $\begingroup$ Nice "hint." ;-))) But +1 for the answer $\endgroup$ – Mark Viola Jul 4 '15 at 17:59
  • $\begingroup$ Yes I believe I wrote too much as a "hint" :D $\endgroup$ – Landon Carter Jul 4 '15 at 17:59

As others have pointed out, you cannot differentiate and think the inequality remains valid. However, you can integrate, retracing your steps.

$$x>e\implies \frac1x<\frac1e\implies \int_e^{\pi}\frac1x\,dx<\int_e^{\pi}\frac1e\,dx\implies \log \pi < \frac{\pi}e\implies \pi^e<e^\pi$$

  • $\begingroup$ Nice proof using integrals. Those who used differentiation could have used implicit differentiation and then the argument would be valid. $\endgroup$ – Shailesh Jul 5 '15 at 2:11

No, your proof is invalid. The idea is good, though: the inequality $$ x^e<e^x $$ is equivalent (for $x>0$) to $e\log x<x$. Consider the function $$ f(x)=x-e\log x $$ defined for $x>0$. Its limits at $0$ and at $\infty$ are both $\infty$.

Next you can consider the derivative: $$ f'(x)=1-\frac{e}{x}=\frac{x-e}{x} $$ which shows the function is decreasing for $0<x<e$ and increasing for $x>e$. The absolute minimum is $$ f(e)=e-e=0 $$ so, for $x>e$, we know that $f(x)>0$.

Thus, for $x\ne e$, $x>e\log x$, that is, $e^x>x^e$. This holds in particular for $\pi\ne e$.


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