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?
 A: HINT.- You have $$\sin x=\sin(\pi-x)$$ Make a change of variables $X=\pi-x$ so you have
$$\frac{\sin X}{X}\le\frac{4x}{\pi^2}$$ Now use the concavity of sinus and your remark on the tangent of $\dfrac{\sin x}{x}$
A: $p_3(x) = \frac{4}{\pi^2}x(\pi-x)$ is an 
interpolation polynomial
for $f(x) = \sin (x)$ at the nodes
$x_0 = 0, x_1= x_2= \frac \pi 2, x_3 = \pi$:
$$
 p_3(0) = f(0) \, , \, 
 p_3(\frac \pi 2) = f(\frac \pi 2) \, , \,
 p_3'(\frac \pi 2) = f'(\frac \pi 2) \, , \,
 p_3(\pi) = f(\pi) \, .
$$
It follows from general 
error estimates for polynomial interpolation
that for $0 \le x \le \pi$
$$
 f(x) - p_3(x) = \frac{f^{(4)}(\xi)}{4!} \prod_{j=0}^4 (x-x_j) 
  = \frac {\sin(\xi)}{24}  x(x-\frac \pi 2)^2(x-\pi) 
$$
for some $\xi \in (0, \pi)$, and therefore
$$
 f(x) - p_3(x) \le 0
$$
with strict inequality for $x \ne 0, \frac \pi 2, \pi$.
We also get a lower bound
$$
 f(x) - p_3(x) \ge -\frac{1}{24} x(x-\frac \pi 2)^2(x-\pi) \ge -\frac{1}{24}
 \ge  -\frac{1}{24} \left( \frac \pi 4 \right)^2 \frac \pi 2 \pi
 \approx - 0.13 \, .
$$

The referenced Wikipedia article contains a proof of error estimate for that case
that all nodes $x_j$ are different, but it holds for "repeated nodes" as well.
Here is a proof for the special case needed above, which also demonstrates the
general idea ($F = f - p_3$):

Let $I \subset \Bbb R$ be an interval and $F:I \to \Bbb R$ have derivatives up to order $4$.
  If $a, b, c \in I$ are distinct and
  $$
 F(a)=F(b)=F(c) = 0 \, , \, F'(b) = 0
$$
  then for all $x \in I$ there exists a $\xi \in I$ such that
  $$
 F(x) = \frac{1}{4!} (x-a)(x-b)^2(x-c) F^{(4)}(\xi) \, .
$$

Proof: There is nothing to show for $x=a, b, c$. From now on let $x_0 \in I$
be fixed with $x_0 \ne a, b, c$.
For $x \in I $ define
$$
P(x) = (x-a)(x-b)^2(x-c) \\
g(x) = P(x_0)F(x) - F(x_0) P(x) \, .
$$
Then
$$
 g(a)=g(b)=g(c)=g(x_0) = 0 \, , \, g'(b) = 0 \, .
$$
Repeated application of Rolle's theorem to $g,g',g'',g'''$ shows
that there is a $\xi \in I$ such that $g^{(4)}(\xi) = 0$.
Then
$$
0 = g^{(4)}(\xi) = P(x_0) \cdot F^{(4)}(\xi) - F(x_0) \cdot P^{(4)}(\xi) \\
 = (x_0-a)(x_0-b)^2(x_0-c) \cdot F^{(4)}(\xi) - F(x_0) \cdot 24
$$
which is what we wanted to show.
A: Hint: show that  $$f(x)=\frac{4}{\pi^2}x(\pi-x)-\sin(x)$$ has in $$x=\frac{\pi}{2}$$ a minimum
A: 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!
A: 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. 
