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?

I'm adding a second answer because the method is very different.
This proof uses the double angle formulas for sine and cosine. From
$$\sin 2x=2\sin x\cos x\qquad\cos2x=2\cos^2x-1$$
we get
$$\tan2x=\frac{\sin2x}{\cos2x}=\frac{2\sin x\cos x}{2\cos^2x-1}>\frac{2x(1-x^2/3)(1-2x^2/3)}{2(1-x^2/3)^2-1},$$
using the bounds $\sin x>x(1-x^2/3)$, $1-2x^2/3<\cos x<1-x^2/3$ derived in the original post. Letting $y=x^2/3$, we have:
\begin{align}\frac{2x(1-y)(1-2y)}{2(1-y)^2-1}>2x&\iff(1-y)(1-2y)>2(1-y)^2-1\\
&\iff1-3y+2y^2>1-4y+2y^2\\
&\iff y>0.
\end{align}
Now $y=x^2/3>0$ for $x\ne0$, and the first step is justified when $2(1-y)^2-1>0$, but $0<\cos 2x<2(1-x^2/3)^2-1$ ensures that this is the case, so we can conclude $\tan2x>2x$ for all $x\in(0,1]$ such that $\cos 2x>0$, which is to say, when $2x\in(0,\pi/2)$ (since $\pi/4<1$).
PS: This answer has been successfully turned into a formal proof.
