Proving $\pi x(1-x)\le \sin \pi x \le 4x(1-x)$ for $x \in [0,1]$ The LHS of the following inequality
 $$\pi x(1-x)\le \sin \pi x \le 4x(1-x), ~~ x \in [0,1]~~~~(1)$$
can be proved by taking $$f(x)=\sin \pi x-\pi x(1-x) \implies f'(x)=\pi(\cos \pi x-1+2x), f''(x)=-\pi^2 \sin \pi x+2$$
Next, $$f'(x)=0 \implies \cos \pi x=1-2x$$ has exactly three real roots: $x=0, 1/2, 1$, it can be checked graphically that $y=\cos \pi x$  and $y=1-2x$ cut each other at these three points only.
As $f''(0)= f''(1)=2 &gt;0$ and $f''(1/2)=2-\pi^2&lt;0.$ Consequently, $$f(x) \ge f_{min}=f(0)\implies \sin \pi x\ge \pi x(1-x).$$

The question here is: How to prove the RHS of (1)?
If you replace $x$ by ${1 \over 2} - x$, the inequality you are trying to prove is
$$ \cos {\pi x} \leq 1 - 4x^2$$
You want to show this for $|x| \leq {1 \over 2}$. By evenness of the functions it suffices to show this on $[0,{1 \over 2}]$. The form of the right-hand side suggests one use  the double angle identity $\cos {\pi x} = 1 - 2\sin^2{\pi x \over 2}$. Making this substitution, we see that we must show
$$ 1 - 2\sin^2{\pi x \over 2} \leq 1 - 4x^2$$
This translates to simply 
$$\sin{\pi x \over 2} \geq \sqrt{2} x$$
This is to be valid on $[0,{1 \over 2}]$. Note that the line $y = \sqrt{2} x$ intersects the graph of $\sin{\pi x \over 2}$ at $(0,0)$ and $({1 \over 2}, {\sqrt{2} \over 2})$, so you need to show that the graph of  $\sin{\pi x \over 2}$ lies above the line $y = \sqrt{2} x$ on $[0, {1 \over 2}]$. This follows from the concavity of $\sin{\pi x \over 2}$ on the interval.
Let $x=\frac12 - \frac{2t}\pi$. Then, the inequality 
$$ \sin \pi x \le 4x(1-x), \&gt;\&gt;\&gt;\&gt;\&gt; x\in [0,1]\tag 1$$

is equivalent to

$$f(t) = \frac{\sin t }t -  \frac{2\sqrt2}\pi \ge 0, \&gt;\&gt;\&gt;\&gt;\&gt; t \in [-\frac\pi4,\frac\pi4]\tag 2$$

Due to the symmetry, consider only the positive domain $t\in[0,\frac\pi4]$. Note that,

$$f(\frac\pi4) = 0,\&gt;\&gt;\&gt;\&gt;\&gt;
f'(t) = \frac{\cos t}{t^2} (t-\tan t)
=-\frac{\cos t}{t^2} \int_0^t \tan^2 udu \le 0$$

which means $f(t) \ge 0 $ for $t \in [0,\frac\pi4]$. Thus, the inequality (2) and its equivalent (1) hold.