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?

With $$f(x)=\sin x-\frac4{\pi^2}x(\pi-x)$$
we have
$$f'(x)=\cos x-\frac4{\pi^2}(\pi-2x),$$
$$f''(x)=-\sin x+\frac8{\pi^2}.$$
The second derivative cancels twice in the range $[0,\pi]$, symmetrically around $\pi/2$, so that the first derivative has two extrema, namely a maximum then a minimum.
Then as
$$f'(0)=1-\frac4{\pi^2}=-f'(\pi)$$ and $$f'\left(\frac\pi2\right)=0$$ the first derivative has three roots corresponding to a sequence minimum/maximum/minimum.
As $$f(0)=f\left(\frac\pi2\right)=f(\pi)=0,$$
the function is never positive. 
