<s>[INST] How prove this inequality $\pi<\frac{\sin{(\pi x)}}{x(1-x)}\le 4$ let $x\in (0,1)$ show that

$$\pi<\dfrac{\sin{(\pi x)}}{x(1-x)}\le 4$$

I idea we know 
$$\sin{x}<x$$
and
$$x(1-x)\le\dfrac{1}{4}$$
But not usefull for this problem 
 [/INST] Let
$$y=\frac{\sin{(\pi x)}}{x(1-x)}$$
then
$$\dfrac{dy}{dx}=\dfrac{{\pi {x(1-x)}\cos(\pi x)-(1-2x)\sin(\pi x)}}{x^2(x-x^2)}$$
For the maximum/minimum values we need to know x that make
$${{\pi {x(1-x)}\cos(\pi x)-(1-2x)\sin(\pi x)}}=0$$
If$\quad$ $x=\frac{1}{2}$, we can easily find $\dfrac{dy}{dx}=0$
If$\quad$ $x\not=\frac{1}{2}$,
$$x(1-x)\pi \cos(\pi x)=(1-2x)\sin{\pi x}$$
so
$$x^2(1-x)^2\pi^2 \cos^2(\pi x)=(1-2x)^2\sin^2{\pi x}$$
so
$${\dfrac{\pi^2 x^2(1-x)^2}{(1-2x)^2}}\cos^2(\pi x)=\sin^2{\pi x}=1-\cos^2(\pi x)$$
so
$${\cos^2(\pi x)}{\biggl[1+\dfrac{\pi^2 x^2(1-x)^2}{(1-2x)^2}\biggr]}=1$$
If we put 
$$g={\cos^2(\pi x)}{\biggl[1+\dfrac{\pi^2 x^2(1-x)^2}{(1-2x)^2}\biggr]-1}$$
then
$$g \gt0, \quad 0\lt x\lt \frac{1}{2}$$
$$g \lt0, \quad \frac {1}{2}\lt x\lt 1$$
$$g =0, \quad if\ x =0\ or\ 1$$
Because x $\in$ (0,1) only x=$\frac12$ will make $\dfrac{dy}{dx}=0$
and 
$$\dfrac{dy}{dx} \gt 0 \quad if \quad 0\lt x \lt \frac12$$
$$\dfrac{dy}{dx} \lt 0 \quad if \quad \frac12 \lt x \lt 1$$
because the denominator of $\dfrac{dy}{dx}$  $$x^2(x-x^2) \gt 0 \quad when \quad x\in (0,1)$$ 
so, y has its maximum at y($\frac12$)=4
and approaches its minimum value as x approaches to 0 or 1 
$$
\lim_{x \to 0}y=\lim_{x \to 1}y = \pi
$$
Therefore $$\pi \lt \frac{\sin{(\pi x)}}{x(1-x)} \le 4$$
</s>