TITLE: Showing $\sin(x) < x$ for all $x>0$ using the mean value theorem
QUESTION [9 upvotes]: I want to show that $\sin(x) < x$ for all $x>0$, using the mean value theorem.
Since the sine is bounded above by $1$, it's obviously true for $x > 1$. Consider $x \in ]0,1]$. Let $f(x)=\sin(x)$. Choose $a=0$ and $x>0$, then there is, according to the mean value theorem, an $x_0$ between $a$ and $x$ with 
$$f'(x_0)=\frac{f(x)-f(a)}{x-a} \Leftrightarrow (\sin(x))'(x_0)= \frac{\sin(x)-\sin(a)}{x} \Leftrightarrow \cos(x_0)=\frac{\sin(x)}{x}$$
Since $1\geq x_0>0 \Rightarrow \cos(x_0) < 1$,
$$\Rightarrow 1 > \cos(x_0)=\frac{\sin(x)}{x} \Rightarrow x  > \sin(x)$$
Is my proof correct?

REPLY [6 votes]: Looks OK.
Generally, the way one proves that if $f'>0$ everywhere then $f$ is increasing is by using the mean value theorem in this same way.  One could apply that idea to $f(x)=x-\sin x$.  Since $f(0)=0$ and $f'(x)=1-\cos x>0$ (for $0<x<2\pi$), one concludes that $f(x)$ increases as $x$ increases from $0$; hence $f(x)>0$ for those values of $x$.  That is a bit different from your way, but either works.