prove $(\sin x)^{-2}-x^{-2}\leq 1-\frac{4}{{\pi}^{2}},x\in(0,\pi/2]$

Question

$(\sin x)^{-2}-x^{-2}\leq 1-\frac{4}{{\pi}^{2}},x\in(0,\pi/2]$

How to deal with this problem? Observing that when $x=\pi/2$, the above inequality becomes equality.

Firstly, denote $f(x)=(\sin x)^{-2}-x^{-2}$ and then take derivative of $f(x)$. We have $$-2(\sin x)^{-3}\cos x+2x^{-3}$$ Next, how to analysis the sign of $f'(x)$?

Any hints are wellcome! Thanks!

I think you meant $\frac{4}{\pi^2}$ and not $\frac{\pi^2}{4}$ — Chavez, Jan 26 '23 at 03:53
what I am thinking is for all $x \in (0, \pi/2]$ $$ \frac{1}{x^2} \geq \frac{4}{\pi^2} $$ Because as denominator increases value of $\frac{1}{x^2}$ decreases and $$ - \frac{1}{x^2} \leq -\frac{4}{\pi^2} \cdots (1) $$ Now take $$ (\sin^2{x})^{-2} = \frac{1}{\sin^2{x}} \leq 1 \cdots(2) $$ Now adding both the equations we get $$ (\sin^2{x})^{-2} - x^{-2} \leq 1 - \frac{4}{\pi^2} $$ — Lucky Chouhan, Jan 26 '23 at 04:17
@LuckyChouhan Please, transform your comment into an answer. — Jean Marie, Jan 26 '23 at 09:34
@monotoneoperator look at https://journalofinequalitiesandapplications.springeropen.com/articles/10.1155/2009/271923 — Miss and Mister cassoulet char, Jan 26 '23 at 13:05

Miss and Mister cassoulet char · Answer 1 · 2023-01-26T14:22:05.867

It's related to :

$$|\frac{1}{\sin(x)}-\frac{1}{x}|\leq 1-\frac{2}{\pi}$$

Called Prestin's inequality

Use the fact :

$$\left|\frac{1}{\sin^{2}(x)}-\frac{1}{x\sin\left(x\right)}\right|-\left(1-\frac{2}{\pi}\right)\geq (\sin x)^{-2}-x^{-2}-\left(1-\frac{4}{\pi^{2}}\right)$$

And :

$$\left|\frac{1}{\sin(x)}-\frac{1}{x}\right|-\left|\sin\left(x\right)\right|\left(1-\frac{2}{\pi}\right)\leq 0$$

Or using Jordan's inequality :

$$\left|\frac{1}{\sin(x)}-\frac{1}{x}\right|-\left|\frac{2}{\pi}x\right|\left(1-\frac{2}{\pi}\right)\leq 0$$

Or using Taylor's series for $x\in(0,1]$ :

$$\left|\frac{1}{x-\frac{1}{6}x^{3}}-\frac{1}{x}\right|-\left|\frac{2}{\pi}x\right|\left(1-\frac{2}{\pi}\right)\leq 0$$

Or :

$$\frac{\left|x\right|\left(\frac{\pi^{2}}{\left|x^{2}-6\right|}-2\pi+4\right)}{\pi^2}\leq 0$$

Now on $x\in[1,\pi/2]$ :

$$\left|\frac{1}{1-1/2(x-\pi/2)^{2}}-\frac{1}{x}\right|-\left|\frac{2}{\pi}x\right|\left(1-\frac{2}{\pi}\right)\leq 0$$

Or :

$$-\frac{\left((2x-\pi)(4\pi x^{3}-8x^{3}-2\pi^{2}x^{2}+4\pi x^{2}+2\pi^{2}x-8\pi x+16x-\pi^{3}+8\pi)\right)}{(\pi^{2}x(4x^{2}-4\pi x+\pi^{2}-8))}\leq 0$$

Thank you very much! – monotone operator Jan 26 '23 at 13:17 — monotone operator, Jan 26 '23 at 13:17
I get your idea, thank you for your patient! – monotone operator Jan 31 '23 at 00:58 — monotone operator, Jan 31 '23 at 00:58

xpaul · Accepted Answer · 2023-03-05T18:51:44.283

4

Let $f(x)=\sin^{-2}x-x^{-2}$. Note $$ f'(x)=\frac{2}{x^3}-\frac{2\cos x}{\sin^3x}=\frac{2(\sin^3x-x^3\cos x)}{x^3\sin^3x}=\frac{g(x)}{x^3\sin^3x}. $$ where $$ g(x)=\sin^3x-x^3\cos x.$$ Now $$ g'(x)=x^3\sin x-3\cos x(x^2-\sin^2x)\ge0, \forall x\in(0,\frac\pi2] $$ which implies that $g(x)$ is increasing and hence $$ g(x)> g(0)=0, \forall x\in(0,\frac\pi2]. $$ This also implies that $f(x)$ is increasing. So $x\in[0,\frac\pi2]$ $$ f(x)\le f(\frac\pi2) $$ or $$(\sin x)^{-2}-x^{-2}\leq 1-\frac{4}{{\pi}^{2}},x\in(0,\pi/2]. $$ $\bf{Update:}$ Actually we don't need $g'(x)\ge0$. To prove $g(x)\ge 0$ for $x\in(0,\pi/2]$, using $$ \sin x\ge x-\frac{x^3}6, \cos x\le 1-\frac{x^2}2+\frac{x^4}{24}, $$ one has \begin{eqnarray} g(x)&=&x^3\bigg[\bigg(\frac{\sin x}{x}\bigg)^3-\cos x\bigg]\\ &\ge&x^3\bigg[\bigg(1-\frac{x^2}6\bigg)^3-\bigg(1-\frac{x^2}2+\frac{x^4}{24}\bigg)\bigg]\\ &=&\frac1{216}x^7(9-x^2)>0, x\in(0,\pi/2]. \end{eqnarray}

edited Mar 05 '23 at 18:51

answered Jan 26 '23 at 15:00

xpaul

44,000

@monotoneoperator You should accept this answer . – Miss and Mister cassoulet char Jan 26 '23 at 15:32
@ErikSatie you are cool guy. Now I got why Empress of Austria is famous for $Si(x)Si(x)$ integral. Because her pet name was "sisi". hahaha – Lucky Chouhan Jan 26 '23 at 17:47
@ErikSatie thank you very much! But I want to know why $g'(x)=x^{3}\sin x-3\cos x(x^{2}-\sin^{2} x)\geq0$. Does it easy to see? – monotone operator Jan 30 '23 at 05:44
@monotoneoperator, See the update. – xpaul Jan 30 '23 at 17:20
@xpaul thank you very much!!!! Your update is very beautiful!! – monotone operator Jan 31 '23 at 00:41

score 1 · Answer 3 · answered Jan 31 '23 at 02:55

1

The Laurent series expansion of $\frac{1}{\sin x} $, valid for $0<|x|< \pi$, is

$$\frac{1}{\sin x} = \frac{1}{x} + \frac{1}{6} x + \frac{7}{360} x^3 + \frac{31}{1520} x^5 + \cdots $$

with all coefficients positive. We conclude that also $\frac{1}{\sin^2 x}$ has a Laurent expansion with positive coefficients valid for $0< |x| < \pi$

$$\frac{1}{\sin^2 x} = \frac{1}{x^2} + \frac{1}{3}+ \frac{1}{15} x^2 + \frac{2}{189} x^4 + \frac{1}{675} x^6 + \cdots$$

From this several inequalities follow.

answered Jan 31 '23 at 02:55

orangeskid

53,909

1

Thank you, your method is so powerful!!! Best wishes!! – monotone operator Jan 31 '23 at 08:44
@monotone operator: Glad it was useful, thank you, best of luck! – orangeskid Jan 31 '23 at 09:05

prove $(\sin x)^{-2}-x^{-2}\leq 1-\frac{4}{{\pi}^{2}},x\in(0,\pi/2]$

3 Answers3