How prove this inequality $\pi<\frac{\sin{(\pi x)}}{x(1-x)}\le 4$

Question

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

Answer 1

You can also translate the variable and prove that $$ \frac{\pi}{4} \leq \frac{\cos(\pi x)}{1-4x^2} \leq 1 $$ for any $x\in [0,1/2]$, since the middle term is an even function. Now, since the Weiestrass product of the cosine function give us: $$ \cos(\pi x) = \prod_{n=0}^{+\infty}\left(1-\frac{4x^2}{(2n+1)^2}\right), $$ we have that the middle term is: $$ f(x)=\frac{\cos(\pi x)}{1-4x^2} = \prod_{n=1}^{+\infty}\left(1-\frac{4x^2}{(2n+1)^2}\right),$$ an infinite product of decreasing functions over $[0,1/2]$, hence a decreasing function, giving us: $$\frac{\pi}{4}=\lim_{x\to 1/2} f(x)\leq f(x)\leq f(0)=1.$$

Answer 2

$f(x)=\dfrac{\sin{\pi x}}{x(1-x)}$

note:$f(x)=f(1-x) \implies f(x) $ are symmetry $x=\dfrac{1}{2}$ when $x \in (0,1)$

now we prove $ f(x) $ will be mono increasing function when $x \in (0,\dfrac{1}{2}]$

$f'(x)= \dfrac{{\pi {x(1-x)}\cos(\pi x)-(1-2x)\sin(\pi x)}}{x^2(1-x)^2}$

$g(x)= \pi x(1-x)\cos(\pi x)-(1-2x)\sin(\pi x) $

$g'(x)=(2 - \pi^2 x(1- x) ) \sin(\pi x)=0$

$sin (\pi x)=0,x_{1}=0$,

$2 - \pi^2 x(1- x)=0, x_2=\dfrac{\pi - \sqrt{\pi^2-8} }{2 \pi}$

verify $g(0)=0 ,g(x_2)>0 \quad \text{(max point of} \ g(x))$

check bound $g(\dfrac{1}{2})=0 \implies g(x)\ge 0 \implies f'(x) >0$ $ f_{max}=f(\dfrac{1}{2})=4,f_{min}=f(0^+)=\displaystyle{\lim_{x \to 0+}}f(x)= \pi $

QED.

Answer 3

When $0<x<1$ one has $${\sin(\pi x)\over x(1-x)}=4\int_0^{\pi/2}\cos t\ \cos(\sigma\>t\bigr)\ dt,$$ where $\sigma:=|2x-1|$.

Inspecting the right hand side one immediately sees that it is a decreasing function of $\sigma\in[0,1]$. Therefore it is maximal $(=4)$ when $\sigma=0$, i.e., $x={1\over2}$, and minimal $(=\pi)$ when $\sigma=1$, i.e., in the limit $x\to0$ or $x\to1$.

Answer 4

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 $$\pi \lt \frac{\sin{(\pi x)}}{x(1-x)} \le 4$$

Answer 5

Hint: Let $x={1+u\over2}$ with $-1\le u\le1$.