TITLE: Prove that $e^\pi+\frac{1}{\pi} < \pi^e+1$
QUESTION [24 upvotes]: 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.

REPLY [12 votes]: 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.