How to prove that $\sin(x)≤\frac{4}{\pi^2}x(\pi-x) $ for all $x\in[0,\pi]$? How to prove the inequality
$$
\sin(x)≤\frac{4}{\pi^2}x(\pi-x)
$$
for all $x\in[0,\pi]$?
As both functions are symmetric to $\frac{\pi}{2}$ it suffices to prove it for $x\in\left[0,\frac{\pi}{2}\right]$. Furthermore one can see that $\frac{4}{\pi^2}(\pi-x)$ is the tangent to $\frac{\sin(x)}{x}$ in the point $\left(\frac{\pi}{2},\frac{2}{\pi}\right)$ so it would be enough to prove that $\frac{d^2}{dx^2}\left[\frac{\sin(x)}{x}\right]≤0$ in $\left[0,\frac{\pi}{2}\right]$. Is this the right way or are there more elegant approaches?

Let $f(x)=\frac{4x(\pi-x)}{4\pi^2}-\sin{x}$.
Since $f(\pi-x)=f(x)$, it's enough to prove our inequality for $x\in\left[0,\frac{\pi}{2}\right]$
$f'(x)=\frac{4(\pi-2x)}{\pi^2}-\cos{x}$ and $f''(x)=\sin{x}-\frac{8}{\pi^2}$, 
which says that $f$ is a concave function on $\left[0,\arcsin\frac{8}{\pi^2}\right]$ 
and $f$ is a convex function on $\left[\arcsin\frac{8}{\pi^2},\frac{\pi}{2}\right]$.
Easy to see that $f\left(\arcsin\frac{8}{\pi^2}\right)>0$, $f(0)=0$, $f\left(\frac{\pi}{2}\right)=0$ and $f'\left(\frac{\pi}{2}\right)=0$.
Thus, graph of $f$ 
located above  AB, 
where $A(0,0)$ and $B\left(\arcsin\frac{8}{\pi^2},f\left(\arcsin\frac{8}{\pi^2}\right)\right)$ on $\left[0,\arcsin\frac{8}{\pi^2}\right]$
and graph of $f$ located above the  $x$-axis on $\left[\arcsin\frac{8}{\pi^2},\frac{\pi}{2}\right]$.
Done!
