Proving $\frac2\pi x \le \sin x \le x$ for $x\in [0,\frac {\pi} 2]$ 
Prove $\frac2\pi x \le \sin x \le x$ for $x\in [0,\frac {\pi} 2]$.

I tried to do this in two ways, I'm not sure about CMVT and I have a problem with the other way.

Using Cauchy's MVT:
RHS: 
$\sin x \le x \implies \frac {\sin x}{x}\le 1$ 
So define: 
$f(x)=\sin x, \ g(x)=x$ then from CMVT: 
$\dfrac {f(\frac {\pi} 2)}{g(\frac {\pi} 2)}=\dfrac {f'(c)} {g'(c)}=\cos c$ 
and from the fact that $c$ is between $0$ and $\pi/2 \implies \cos c \le 1$.
LHS: In the same manner but here I run into some trouble:
$\frac2\pi x \le \sin x\implies \frac {2x}{\pi\sin x}\le 1$
So: 
$\dfrac {f(\frac {\pi} 2)}{g(\frac {\pi} 2)}=\dfrac {f'(c)} {g'(c)}\implies\frac {1}{\sin {\frac {\pi}{2}}}=\frac {2}{\pi \cos c}$
 Here actually 
$\frac {1}{\sin {\frac {\pi}{2}}}=1$ so it's also $\le 1$
Is it correct to use CMVT like this ?

The other way: 
We want to show: $f(x)=\sin x - x < 0$ and $g(x)=\frac {2x}{\pi}-sinx <0 $ by deriving both it's easy to show that the inequality stands for $f$ but for $g$ it isn't so obvious that $g'(x)=\frac {2}{\pi}-\cos x$ is negative. In fact for $x=\frac {\pi} 2$ it's positive. Please help figure this out.

This is the same The sine inequality $\frac2\pi x \le \sin x \le x$ for $0<x<\frac\pi2$ but all the answers there are partial or hints and I want to avoid convexity.
Note: I can't use integrals.

For any $x \in (0,\frac{\pi}{2})$, consider the expression
$$\frac{\sin x - \sin 0}{x - 0} - \frac{\sin\frac{\pi}{2} - \sin x}{\frac{\pi}{2}- x}
= \frac{\sin x}{x} - \frac{1-\sin x}{\frac{\pi}{2} - x}\tag{*1}
$$
Apply MVT on for the first term on $[0,x]$ and the second term on $[x,\frac{\pi}{2}]$, we can find two numbers $y, z$ such that
$$0 < y < x < z < \frac{\pi}{2}\quad\text{ and }\quad  \frac{\sin x}{x} = \cos y \;\land\; \frac{1-\sin x}{\frac{\pi}{2} - x} = \cos z$$
Since $\cos t$ is strictly decreasing on $[0,\frac{\pi}{2}]$, we have $\cos y > \cos z$ and hence
$$\begin{align}\frac{\sin x}{x} - \frac{1-\sin x}{\frac{\pi}{2} - x} > 0
&\iff
\left(\frac{\pi}{2} - x \right)\sin x - x \left(1 - \sin x\right) >  0\\
&\iff
\sin x > \frac{2x}{\pi}\end{align}$$
You may wonder how I arrive the expression in $(*1)$. Geometrically,


*

*$\displaystyle\;\frac{\sin x}{x}\;$ is the slope of $\sin x$ over $[0,x]$.

*$\displaystyle\;\frac{1-\sin x}{\frac{\pi}{2} - x}\;$ is the slope over $[x,\frac{\pi}{2}]$. 


This proof works because the slope $\cos x$ is decreasing on $[0,\frac{\pi}{2}]$. 
This is sort of equivalent to $\sin''(x) = -\sin x < 0$. In certain sense, this is really a proof with convexity hiding under the carpet.
