TITLE: Algebraic proof of $\tan x>x$
QUESTION [5 upvotes]: I'm looking for a non-calculus proof of the statement that $\tan x>x$ on $(0,\pi/2)$, meaning "not using derivatives or integrals." (The calculus proof: if $f(x)=\tan x-x$ then $f'(x)=\sec^2 x-1>0$ so $f$ is increasing, and $f(0)=0$.) $\tan x$ is defined to be $\frac{\sin x}{\cos x}$ where these are defined by their infinite series. What I have so far:
$$|z|\le1\implies\left|\sum_{n=4}^\infty\frac{z^n}{n!}\right|<\sum_{n=0}^\infty\frac{|z|^4}{4!\,5^n}=\frac{5|z|^4}{4\cdot 4!}$$
$$\left|\sin x-\Big(x-\frac{x^3}6\Big)\right|=\Im\left[\sum_{n=4}^\infty\frac{(ix)^n}{n!}\right]<\frac{5x^4}{4\cdot 4!}<\frac{x^3}6$$
$$\left|\cos x-\Big(1-\frac{x^2}2\Big)\right|=\Re\left[\sum_{n=4}^\infty\frac{(ix)^n}{n!}\right]<\frac{5x^4}{4\cdot 4!}<\frac{x^2}6$$
Thus $\sin x>x-\frac{x^3}3$ and $\cos x<1-\frac{x^2}3$, so $\tan x>x$. However, this only covers the region $x\le1$, and I still need to bound $\tan x$ on $(1,\pi/2)$. My best approximation to $\pi$ is the very crude $2<\pi<4$, derived by combining the above bounds with the double angle formulas (note that $\pi$ is defined as the smallest positive root of $\sin x$), so I can't quite finish the proof with a bound like $\sin x>1/\sqrt 2$, $\cos x\le\pi/2-x$ (assuming now $x\ge1\ge\pi/4$) because the bound is too tight. Any ideas?

REPLY [5 votes]: Here is a not-so-original geometric proof.

In unit circle $F$, draw $\Delta ABC$ with $C$ at the center of $F$ and  $\overline{AB}$ tangent to $F$ at $A$. Let $D$ be a point on $\overline{BC}$ and $E$ be a point on $\overline{AB}$ such that $\overline{DE}$ is tangent to $F$ at $D$. Furthermore, let $\theta=\measuredangle ACB$.
It is clear that $\tan\theta = \lvert\overline{AB}\rvert$ because $\measuredangle CAB = \pi/2$ due to tangency. As well, $\tan\theta=\lvert\overline{AE}\rvert+\lvert\overline{EB}\rvert$, but $\lvert\overline{ED}\rvert\lt\lvert\overline{EB}\rvert$ because $\lvert\overline{EB}\rvert$ is the hypotenuse of $\Delta BDE$. Thus we have
$$\tan\theta\gt\lvert\overline{ED}\rvert+\lvert\overline{EA}\rvert$$
From the diagram we see that sector ACD $\subset$ quadrilateral ACDE, and because both are convex sets, the perimeter of sector ACD $\lt$ perimeter ACDE or
$$\lvert\overline{AC}\rvert+\lvert\overline{CD}\rvert+\lvert arc\:DA \rvert \lt \lvert\overline{AC}\rvert+\lvert\overline{CD}\rvert + \lvert\overline{DE}\rvert+\lvert\overline{EA}\rvert$$
$$\lvert arc\:DA \rvert \lt  \lvert\overline{DE}\rvert+\lvert\overline{EA}\rvert$$
Finally we have
$$\theta=\lvert arc\:DA \rvert \lt  \lvert\overline{DE}\rvert+\lvert\overline{EA}\rvert\lt\lvert\overline{EA}\rvert+\lvert\overline{EB}\rvert=\tan\theta$$