Algebraic proof of $\tan x>x$ 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?

Here is a sketch of what you might be looking for:
Showing $\tan x > x$ is equivalent to showing $\sin x - x \cos x > 0$, since $\cos x > 0$ on $(0,\pi/2$).
The series for $\sin x - x \cos x$ is $\displaystyle\sum_{j=1}^{\infty} \dfrac{(2j)x^{2j+1}}{(2j+1)!} = x^3/3 - x^5/30 + x^7/840 - x^9/45360 \ldots$
Group the terms in pairs: $(x^3/3 - x^5/30) + (x^7/840 - x^9/45360) + \ldots$.  If $0 < x < \sqrt{10}$, the first difference is positive.  The ratio of the terms in each difference is decreasing, so if the first difference is positive, all the rest are too, and the sum is positive.  So $\sin x - x \cos x > 0$ on $0 < x < \sqrt{10}$, which gives you quite a bit of leeway since $\sqrt{10} > \pi/2$.  (The first positive solution to $\sin x - x \cos x = 0$ happens at $x \approx 4.493$ according to WolframAlpha.)
