Prove that:

$$e^\pi+\frac{1}{\pi}< \pi^{e}+1$$

Using Wolfram Alpha $\pi e^{\pi}+1 \approx 73.698\ldots$ and $\pi(\pi^{e}+1) \approx 73.699\ldots$

Can this inequality be proven without brute-force estimations (anything of the sort $e\approx 2.7182...$ or $\pi \approx 3.1415...$)? I've just seen this and I remembered I've seen the question asked here in an older paper, but I don't remember the details.

Note that this is sharper because it can be written as:


I've tried, but none of the methods in the linked question (which study the function $x^\frac{1}{x}$) can be applied here.

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    $\begingroup$ What about the function $f(x)=e^x-x^e$? $\endgroup$ Feb 24, 2020 at 20:50
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    $\begingroup$ @WeierstraßRamirez, That is essentially the same as studying $x^{\frac{1}{x}}$. It has two critical points at $1$ (maximum) and $e$ (minimum). I think it's only enough to show $e^{\pi} > \pi^e$. Or is there something I'm missing? $\endgroup$
    – LHF
    Feb 24, 2020 at 20:58
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    $\begingroup$ Maybe useful the next link! $\endgroup$
    – user0410
    Feb 24, 2020 at 21:38
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    $\begingroup$ What exactly are "estimations"? What exactly may the solution involve? $\endgroup$
    – joriki
    Feb 24, 2020 at 22:19
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    $\begingroup$ Maybe interesting. Another sharp bound for the expression $e^\pi-\pi^e$ is given by $$ e^\pi-\pi^e \approx \frac{1}{6}\,\sqrt [3]{75+7\,\sqrt {449}}-\,{\frac {2}{\sqrt [3]{75+7\,\sqrt {449}}}}<1-\frac{1}{\pi} $$ In fact, I changed the real root of the polynomial $x^3+x-1$, slightly! $\endgroup$
    – Amin235
    Feb 26, 2020 at 18:13

1 Answer 1


From the continued fraction expansion of $\pi$, we have

$$ \frac{333}{106}\lt\frac{103993}{33102}\lt\pi\lt\frac{355}{113}\;. $$

There are various ways of proving these inequalities without using decimal approximations:

In the case of $\mathrm e$, the continued fraction expansion is regular and can be systematically derived (see e.g. A Short Proof of the Simple Continued Fraction Expansion of e by Henry Cohn, The American Mathematical Monthly, $113(1)$, $57$$62$, The Simple Continued Fraction Expansion of e by C. D. Olds, The American Mathematical Monthly, $77(9)$, $968$$974$, or Continued fraction for e at Topological Musings); it yields

$$ \frac{1264}{465}\lt\mathrm e\lt\frac{1457}{536}\;. $$

Thus it suffices to show that

$$ \left(\frac{1457}{536}\right)^\frac{355}{113}+\frac1{\frac{333}{106}}\lt\left(\frac{103993}{33102}\right)^\frac{1264}{465} + 1\;, $$


$$ \left(\frac{1457}{536}\right)^\frac{355}{113}\lt\left(\frac{103993}{33102}\right)^\frac{1264}{465} + \frac{227}{333}\;. $$

Since both sides contain fractional exponents, it’s hard to compare them directly; but we can find a fraction that lies between them and compare them to it separately. Among the suitable fractions, the one with the lowest denominator is $\frac{4767}{206}$. The rational inequalities

$$ \left(\frac{1457}{536}\right)^{355}\lt\left(\frac{4767}{206}\right)^{113} $$


$$ \left(\frac{4767}{206}-\frac{227}{333}\right)^{465}\lt\left(\frac{103993}{33102}\right)^{1264} $$

are readily checked with integer arithmetic, and thus with

$$ \left(\frac{1457}{536}\right)^\frac{355}{113}\lt\frac{4767}{206}\lt\left(\frac{103993}{33102}\right)^\frac{1264}{465} + \frac{227}{333} $$

the result follows.

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    $\begingroup$ I appreciate the great work and references. Given the sharpness of the inequality I doubt a better approach can be found, but I'll wait a little more time (a day or two) before accepting the answer. $\endgroup$
    – LHF
    Feb 25, 2020 at 13:19
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    $\begingroup$ With the utmost respect for you, I appreciate your intervention to this post but how to deny that underlying calculations of it are far more complicated than simply calculating both sides of the inequality. Impossible that I put you a downvote but in no way, due to the type of problem posed, would you put an upvote. Best regards. $\endgroup$
    – Piquito
    Feb 25, 2020 at 14:07
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    $\begingroup$ @Piquito: I think you'd need to explicate your concept of "simply calculating". The question asked for a proof, so you'd need to be able to control the errors in this "simple calculation". Your own answer works with approximations ($\approx$) without specifying any error bounds for them; thus it doesn't constitute a proof. Moreover, the question explicitly asked for answers that don't use this sort of approximation. $\endgroup$
    – joriki
    Feb 25, 2020 at 14:11
  • $\begingroup$ Kind of response to problems with parallel thinking. In this case, calculate everything you want out of the problem and apply it here. My best wishes for you. $\endgroup$
    – Piquito
    Feb 25, 2020 at 15:51
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    $\begingroup$ @joriki, it seems like an epidemy of mysterious senseless downvotes on MSE nowadays. $\endgroup$
    – LHF
    Feb 25, 2020 at 17:14

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