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

AI Thread Summary
The discussion centers on proving the inequality $\dfrac{\pi^2-x^2}{\pi^2+x^2}>\left(\dfrac{\sin x}{x}\right)^2$ for the interval $0<x<\dfrac{\pi}{2}$. Participants share their solutions, with one user expressing satisfaction after solving the challenge. They reference the infinite product representation of $\frac{\sin x}{x}$ and derive inequalities to support their proof. The conversation highlights the use of mathematical properties to establish the inequality, with multiple users confirming similar methods. The challenge fosters engagement and collaboration among participants in the mathematical community.
anemone
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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...
 
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anemone said:
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

$\chi$ $\sigma$
 
Last edited:
Hi chisigma! Thanks for your good solution! And I solved it using the similar method.:)
 
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