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 $f''(0)= f''(1)=2 >0$ and $f''(1/2)=2-\pi^2<0.$ 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)?
 A: 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.
A: Lemma Suppose that there are two functions $ f $ and $ g $ defined on the closed interval $ [a,b] $. Suppose that they are both continuous in $ (a,b) $, are right continuous at $ a $, left continuous at $ b $, and $ f(a) \le g(a) $. Suppose also that $ f $ and $ g $ are both differentiable on $ (a,b) $ and $ f'(c) \le g'(c) $ for all $ c $ in $ (a,b) $. Then $ f(c) \le g(c) $ for all $ c $ in $ [a,b] $.
Proof Suppose that the lemma is not true. Since $ f(a) \le g(a) $, there must be some $ c $ in $ (a,b] $ such that $ f(c) \gt g(c) $. Then $ \frac{f(c)-f(a)}{c-a} \gt \frac{g(c)-g(a)}{c-a} $ for some c in $ (a,b] $. According to the mean value theorem, there is some $ c^* $ in $ (a,c) $ such that $ f'(c^*) = \frac{f(c)-f(a)}{c-a} \gt \frac{g(c)-g(a)}{c-a} = g'(c^*) $, which is a contradiction. $ \square $
Now let $ f(x) = \pi x(1-x) $, $ g(x) = \sin (\pi x) $ and $ h(x) = 4x(1-x) $. We observe that $ f(0) = g(0) = 0 $, $ f'(0) = g'(0) = \pi $, $ f''(0) = -2 \pi \lt 0 = g''(0) $. Since $ f'''(x) = 0 $, $ g'''(x) = -(\pi ^3) \cos (\pi x) $(, let us agree that $ \cos (c) \le 0 $ for all $ c $ in $ [\frac \pi 2,π] $,) $ f(x) \le g(x) $ for all $ x $ in $ [\frac 1 2,1] $.
Using similar method, we can prove that $ g(x) \le h(x) $ for all $ x $ in $ [0,\frac 1 2] $. Since $ f $, $ g $ and $ h $ are symmetric, we can conclude that $ f(x) \le g(x) \le h(x) $ for all $ x $ in $ [0,1] $. $ \square $
A: Let $$g(x)=\sin \pi x-4x+4x^2 \implies g'(x)=\pi \cos \pi x -4+8x,~~ g''(x)=-\pi^2 \sin \pi x+8.$$
$$g'(x)=0 \implies \frac{\pi}{4}\cos \pi x=1-2x.$$
Let $$h(x)=\frac{\pi}{4}\cos \pi x- (1-2x) \implies h(1/2)=0, h(0)=\pi/4-1<0, h(1)=1-\pi/4$$
So $x=1/2$ is one root of () and two more roots $x_1,x_2$ are in $(0,1/2)$ and $(1/2,1)$. Graphically by sketching $y=\frac{\pi \cos \pi x}{4}$ and $y=1-2x$ on can check that these two will cut only at three points. As $f''(1/2)=-\pi^2+8$, so $f_{max}=f(1/2)=0.$ Hence
$$f(x)\le f(1/2) \implies \sin \pi x \le 4x(1-x),~~x \in [0,1]$$
A: Have you tried to use the mean value theorem for the function $f(x)=-\dfrac{1}{\pi}cos\pi x$, $x\in [0,1]$ ? Just a thought...
A: Since my answer got deleted I'm putting my corrected answer here instead.
Let $f(x) = 4x(1-x)-\sin\pi x$.  Since $f$ is symmetric by reflection about $x=\frac12$ it suffices to prove $f(x)\geq 0$ on $[0,\frac12]$.
Let's compute the first four derivatives:
$f'(x) = 4-8x-\pi\cos\pi x$, $f''(x) = -8+\pi^2\sin\pi x$, $f^{(3)}(x) = \pi^3\cos\pi x$, and $f^{(4)} = -\pi^4\sin \pi x$.
Let $x_0 = \frac{8}{\pi^3}$.  Then for $x\in(0,x_0)$ we have $f''(x) \geq -8 + \pi^3 x_0 = 0$ since $\sin\theta\leq\theta$ and the function is increasing on $[0,\frac\pi2]$.  It follows that $f$ is concave down on $[0,x_0]$. Adding $f(0) = 0$, $f(x_0) \approx 0.041 > 0$ we conclude that $f$ is non-negative on this sub-interval.
Let $a=\frac12$. To show $f$ is non-negative on the interval $[x_0,a]$ we shall use Taylor expansion.  Evaluating the derivatives above we have:
$f(a) = f'(a) = f^{(3)}(a) = 0$, $f''(a) = \pi^2-8$.
It follows that on $[x_0,\frac12]$ we have $f(x) = \frac{\pi^2-8}{2}\left(x-\frac12\right)^2-\frac{\pi^4}{24}\sin(\pi y)\left(x-\frac12\right)^4$ for some $x<y<\frac12$, and hence
$f(x) \geq \left(x-\frac12\right)^2\left[\frac{\pi^2-8}{2}-\frac{\pi^4}{24}\left(x-\frac12\right)^2\right]$
on $[x_0,\frac12]$.  Since the parenthetical is monotone, to show that $f$ is non-negative there it suffices to evaluate
$\frac{\pi^2-8}{2}-\frac{\pi^4}{24}\left(x_0-\frac12\right)^2 \approx 0.697>0$.
A: Let $x=\frac12 - \frac{2t}\pi$. Then, the inequality 
$$ \sin \pi x \le 4x(1-x), \>\>\>\>\> x\in [0,1]\tag 1$$
is equivalent to
$$f(t) = \frac{\sin t }t -  \frac{2\sqrt2}\pi \ge 0, \>\>\>\>\> t \in [-\frac\pi4,\frac\pi4]\tag 2$$
Due to the symmetry, consider only the positive domain $t\in[0,\frac\pi4]$. Note that,
$$f(\frac\pi4) = 0,\>\>\>\>\>
f'(t) = \frac{\cos t}{t^2} (t-\tan t)
=-\frac{\cos t}{t^2} \int_0^t \tan^2 udu \le 0$$
which means $f(t) \ge 0 $ for $t \in [0,\frac\pi4]$. Thus, the inequality (2) and its equivalent (1) hold.
