Prove Inequality for $0<x<\dfrac{\pi}{2}$: Math Challenge

[anemone]

For $0<x<\dfrac{\pi}{2}$, prove that $\dfrac{\pi^2-x^2}{\pi^2+x^2}>\left(\dfrac{\sin x}{x}\right)^2$.

I personally find this challenge very intriguing and I solved it, and feel good about it, hehehe...

 

[chisigma]

For $0<x<\dfrac{\pi}{2}$, prove that $\dfrac{\pi^2-x^2}{\pi^2+x^2}>\left(\dfrac{\sin x}{x}\right)^2$.

I personally find this challenge very intriguing and I solved it, and feel good about it, hehehe...

[sp]Rememering that is...

$\displaystyle \frac{\sin x}{x} = \prod_{n=1}^{\infty} (1 - \frac{x^{2}}{n^{2}\ \pi^{2}})\ (1)$

... You derive that...

$\displaystyle (\frac{\sin x}{x})^{2} < (1 - \frac{x^{2}}{\pi^{2}})^{2} < \frac{(1 - \frac{x^{2}}{\pi^{2}})}{(1 + \frac{x^{2}}{\pi^{2}})}\ (2)$

Tha last inequality is justified by tha fact that...

$\displaystyle (1 - \frac{x^{2}}{\pi^{2}})\ (1 + \frac{x^{2}}{\pi^{2}}) < 1\ (3)$[/sp]

Kind regards

$\chi$ $\sigma$

 

[anemone]

Hi chisigma! Thanks for your good solution! And I solved it using the similar method.:)