How do I prove that $3<\pi<4$? Let's not invoke the polynomial expansion of $\arctan$ function.
I remember I saw somewhere here a very simple proof showing that $3<\pi<4$ but I don't remember where I saw it.. (I remember that this proof is also in Wikipedia)
How do I prove this inequality?
My definition for $\pi$ is the twice the first positive real number such that $\cos x= 0$ where $\cos x = \frac{e^{ix} + e^{-ix}}{2}$

I assume that $x\to e^x$ is the unique differentiable function that is its own derivative and maps $0\mapsto 1$.
From this (especially using uniqueness) one quickly establishes $e^{x+y}=e^xe^y$, $e^{\bar z}=\overline{e^z}$ etc.
With the definitions $\cos x=\frac{e^{ix}+e^{-ix}}{2}$, $\sin x=\frac{e^{ix}-e^{-ix}}{2i}$ we then see that $\cos $ and $\sin$ map reals to reals and that $\cos^2x+\sin^2x=1$, $\sin'=\cos$, $\cos'=-\sin$ etc.
Claim 1: There is no $a$ such that $\sin x>\frac12$ for all $x\in [a,a+4]$.
Proof: Otherwise, $2\ge\cos (a+2)-\cos a=\int_a^{a+4}\sin x\,\mathrm dx>2$. $_\square$
Claim 2: There exists $a>0$ such that $\cos a=0$.
Proof: Otherwise we would have $\cos x>0$ for all $x>0$, hence $\sin x$ strictly increasing towards some limit $s:=\lim_{x\to+\infty}\sin x\in(0,1]$. Then $\lim_{x\to\infty}\cos x=\sqrt{1-s^2}=0$ and hence $s=1$. On the other hand, $s\le \frac12$ by claim 1. $_\square$.
By continuity, there exists a minimal positive real $p$ such that $\cos p=0$.
By the IVT, $-\sin \xi =\frac{0-1}{p}$, i.e. $\sin\xi=\frac1p$ for some $\xi\in(0,p)$. This already shows $$p\ge1.$$
Consider the function $f(x)=\cos x-1+\frac12x^2$. We have $f'(x)=-\sin x+x$, $f''(x)=1-\cos x\ge 0$ for all $x$; hence $f'(x)\ge f'(0)$ for all $x\ge 0$, i.e. $\sin x\le x$ for all $x\ge 0$; hence $f(x)\ge f(0)=0$ for all $x\ge0$. We conclude $\cos 1\ge \frac12$. In fact the iniequality is strict, i.e. $\cos1>\frac12$ as otherwise we'd have $f''(x)=0$ for all $x\in[0,1]$. From $\cos'=-\sin\ge -1$ we then see that $\cos (1+x)>\frac12-x$ for $x> 0$. Thus
$$p> \frac32.$$
Consider $g(x)=\cos x-1+\frac12x^2-\frac1{24}x^4$. Then $g'(x)=-\sin x+x-\frac16x^3$, $g''(x)=-\cos x-1-\frac12x^2=-f(x)\le 0$ for $x\ge 0$.
Hence $g'(x)\le g'(0)=0$ for all $x\ge 0$ and finally $g(x)\le g(0)=0$ for all $x\ge 0$.
We conclude $0\ge g(2)=\cos 2-1+\frac42-\frac{16}{24}$, i.e. $\cos 2\le-\frac13$ and hence
$$ p<2.$$
