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

Question

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.

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:

• The accepted answer to How to find continued fraction of pi shows how to find the continued fraction expansion without using decimal approximations as inputs.
• This answer to Is there an integral that proves $\pi > 333/106$? provides integrals with positive integrands that evaluate to the differences in these inequalities.
• You can sum a couple of terms e.g. of the Bailey–Borwein–Plouffe formula for $\pi$, bound the remainder with a geometric series and compare the resulting fractions to the bounds above.

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.