Proving that $\sin x > \frac{(\pi^{2}-x^{2})x}{\pi^{2}+x^{2}}$ Proving that $$\sin x > \frac{(\pi^{2}-x^{2})x}{\pi^{2}+x^{2}}, \qquad\forall x>\pi$$

Hint: say $$f(x) = \sin(x) - \frac{x(\pi^2 - x^2)}{\pi^2+x^2}$$ now $$f'(x) = \cos(x) + \frac{x^4 + 4\pi^2x^2 - \pi^4}{x^4 + 2\pi^2x^2 + \pi^4} = \cos x + 1 + \frac{2\pi^2\left ( x^2 - \pi^2 \right )}{\left ( x^2 + \pi^2 \right )^2}$$ then it is clear that $f'(x) > 0, \forall x > \pi$ and $f(\pi) = 0$ hence we can conclude $f(x) > 0, \forall x > \pi$.
