# Prove that $e^\pi+\frac{1}{\pi} < \pi^e+1$

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:

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

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

• What about the function $f(x)=e^x-x^e$? Feb 24, 2020 at 20:50
• @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?
– LHF
Feb 24, 2020 at 20:58
• Maybe useful the next link! Feb 24, 2020 at 21:38
• What exactly are "estimations"? What exactly may the solution involve? Feb 24, 2020 at 22:19
• 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! Feb 26, 2020 at 18:13

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

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

or

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

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

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

and

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

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.

• 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.
– LHF
Feb 25, 2020 at 13:19
• 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. Feb 25, 2020 at 14:07
• @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. Feb 25, 2020 at 14:11
• 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. Feb 25, 2020 at 15:51
• @joriki, it seems like an epidemy of mysterious senseless downvotes on MSE nowadays.
– LHF
Feb 25, 2020 at 17:14