Non-Numerical proof of $e<\pi$ This is a "coffee-time-style" problem ( to have a taste of this style, you may like to browse the book https://www.amazon.com/Art-Mathematics-Coffee-Time-Memphis/dp/0521693950) interpreted from an anonymous problem once on the interactive whiteboard at my department, namely how to prove $e<\pi$ without much numerical computation like Taylor expansion or so. I once tried to use some "intrinsic connection" between $e$ and $\pi$ like $\sqrt{\pi}=\int_{-\infty}^{+\infty}e^{-x^2}\mathrm{d}x$ ( you can even find it in this movie http://www.imdb.com/title/tt4481414/ for testing children) and one possible way of reducing the problem is in the next paragraph. However it seems to be not that easy, any suggestion or new ideas? 
A stronger version of this question is : can we construct an explicit function $f(x)$ on $\mathbb{R}$ so that $f(x)\leq e^{-x^2}$ for all $x\in\mathbb{R}$ with $f(x)< e^{-x^2}$ on an open interval, and that $\int_{-\infty}^{\infty}f(x)\mathrm{d}x=\sqrt{e}$ ? We know from standard measure theory that there are $\beth_2$ such kind of Lebesgue-integrable functions, but this is the thing: how simple and explicit can what we're looking for be? Examples of very simple and explicit functions include but are not limited to piecewise elementary functions (https://en.wikipedia.org/wiki/Elementary_function). Unfortunately a function $f(x)$ defined piecewisely by  $$f(x)|_{(-1,1)}=e^{-|x|^r}\ \text{where}\ r\in\mathbb{Q}\cap(-\infty,2)\ \text{or}\ \mathbb{Q}\cap (-\infty,2]\ \text{respectively}$$  and $$f(x)|_{(-\infty,-1]\cup[1,\infty)}=e^{-|x|^s}\ \text{where}\ s\in\mathbb{Q}\cap [2,\infty)\ \text{or}\ \mathbb{Q}\cap(2,\infty)$$  would NOT satisfy $\int_{-\infty}^{\infty}f(x)\mathrm{d}x=\sqrt{e}$, if the values of the Gamma  function at rational points are linearly (or even algebraically) independent with $\sqrt{e}$ (https://en.wikipedia.org/wiki/Particular_values_of_the_gamma_function). The question is then how to move on from this first failure to search other explicit functions.
I am aware that it is probably hard to ask such a question as solid as "can we prove that CH is independent from ZFC"; after all, one can argue that any numerical inequality essentially also comes from some intrinsic inequality and hence not numerical at all. However one might try to ask in a relatively sloppy way: is there something that is at least seemingly simpler or less numerical, if not completely non-numerical ?
 A: Incoming overkilll! The error function has both a simple series representation and a simple continued fraction representation, allowing to produce nice algebraic approximations for the Mills ratio.
This answer of mine on MO proves the inequality
$$ \sqrt{\frac{\pi}{2}}\,e^{k^2/2}\, \text{Erfc}\left(\tfrac{k}{\sqrt{2}}\right) \geq \frac{2}{k+\sqrt{k^2+4}} \tag{1}$$
by only using Fubini's theorem and the elementary (convexity) inequality $\frac{2}{\pi}x<\sin(x)<x$ for $x\in\left(0,\frac{\pi}{2}\right)$. By considering $(1)$ at $k=1$ we have
$$ \sqrt{e}\left(\sqrt{\frac{\pi}{2}}-\sum_{n\geq 0}\frac{(-1)^n}{2^n(2n+1)n!}\right)\geq \frac{1}{\varphi} \tag{2}$$
relating $e,\pi$ and the golden ratio $\varphi$ (and allowing to prove $\pi>e$, of course). 
This is pretty much in the spirit of the "natural" relation between $\pi$ and $e$ given by $\pi=\Gamma\left(\frac{1}{2}\right)^2$ and $\Gamma(s)=\int_{0}^{+\infty}x^{s-1} e^{-x}\,dx.$
A: By the Cauchy-Schwarz inequality together with the equality condition, we have
$$ 1 = \left( \int_{0}^{1} dx \right)^2 < \left( \int_{0}^{1} \frac{dx}{1+x^2} \right)\left( \int_{0}^{1} (1+x^2) \, dx \right) = \frac{\pi}{3}. $$
Now utilizing the inequality $e^{-x} \geq 1 - x$ which is true for all $x \in \mathbb{R}$,
$$ \frac{1}{e} = \int_{0}^{1} (1-x)e^{-x} \, dx \geq \int_{0}^{1} (1-x)^2 \, dx = \frac{1}{3}. $$
Combining two inequality yields $\pi > e$.
A: There's not a lot of computation here, just some simple arithmetic and easy upper bounds (plus one important identity):
$$\begin{align}
e^2&=1+2+{2^2\over2}+{2^3\over6}+{2^4\over24}+{2^5\over120}+\cdots\\
&=1+2+2+{4\over3}+{2\over3}+{32\over120}\left(1+{2\over6}+{4\over42}+\cdots \right)\\
&\lt7+{1\over2}\left(1+{1\over2}+{1\over4}+\cdots\right)\\
&=8\\
&\lt6+{3\over2}+{2\over3}\\
&\lt6+{6\over4}+{6\over9}+{6\over16}+{6\over25}+\cdots\\
&=6\sum_{n=1}^\infty{1\over n^2}\\
&=\pi^2
\end{align}$$
Added later: Here's an alternative, which uses the "easy" geometric inequality $\pi\gt3$ (comparing the circumference of a circle to the perimeter of an inscribed hexagon, as in José Carlos Santos's answer) and a small amount of computation:
$$\ln\pi\gt\ln3=-\ln(1/3)=-\int_1^{1/3}{dx\over x}=\int_0^{2/3}{du\over1-u}=\int_0^{2/3}\left(1+u+u^2+\cdots\right)du\\
=\left(2\over3\right)+{1\over2}\left(2\over3\right)^2+{1\over3}\left(2\over3\right)^3+{1\over4}\left(2\over3\right)^4+\cdots\\
\gt{2\over3}+{2\over9}+{8\over81}+{4\over81}={54+18+12\over81}={84\over81}\gt1=\ln e$$
A: Here's my contribution:  We need to show that $\ln \pi >1$.  So we need to show that $\int_1^\pi \frac{1}{x} \; dx >1.$  The graph of $y=\frac{1}{x}$ is concave up, so any tangent line lies below the curve.  Find the tangent line half way through the interval:
$$y = \frac{4}{(\pi+1)^2}(\pi +1 -x).$$
The area under that line and between $x=1$ and $x=\pi$ is less than the area represented by the integral.  So we need to show the area of this trapezoid is greater than $1$.  After some algebra that area turns out to be
$$2\left(\frac{\pi-1}{\pi+1}\right).$$
So we need to show this expression is greater than $1$.  I finally have to stoop to "numerics" and use the fact that $\pi$ is greater than $3$.  Either note that $(x-1)/(x+1)$ is an increasing function and hence the last expression is greater than $2(3-1)(3+1) = 1.$  Or calculate as follows:
$$2\left(\frac{\pi-1}{\pi+1}\right) = 2 - \frac{4}{\pi+1} > 2-\frac{4}{4} = 1.$$
A: Just a suggestion.
$$\int_{-\infty}^\infty\frac{\cos x}{x^2+1}\operatorname d\!x=\frac\pi e$$
If you can prove that the above integral is $>1$ you know $\pi > e$.
A: If we define $e$ as $\lim_{n\to\infty}\left(1+\frac1n\right)^n$, then $e<3$ because, for each $n\in\mathbb{N}$,\begin{align}\left(1+\frac1n\right)^n&\leqslant1+1+\frac1{2!}+\cdots+\frac1{n!}\\&<1+1+\frac12+\frac1{2^2}+\cdots+\frac1{2^{n-1}}\\&<3.\end{align}On the other hand, $2\pi$ is greater than the perimeter of a regular hexagon inscribed in a circle with radius $1$, which is $6$. Therefore, $\pi>3>e$.
