Proving that $\sin x > \frac{(\pi^{2}-x^{2})x}{\pi^{2}+x^{2}}$ Proving that $$\sin x > \frac{(\pi^{2}-x^{2})x}{\pi^{2}+x^{2}}, \qquad\forall x>\pi$$
 A: 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 $f(x) > 0, \forall x > \pi$.
A: With the substitution: $x=t+\pi$ the inequality can be rewritten as
$$
{\sin t\over t}<1+{\pi t\over 2\pi^2+2\pi t+t^2},\quad\hbox{for $t>0$},
$$
which is obviously true.
A: We just need to prove the inequality:
$$\frac{\sin x}{x}\geq \frac{\pi^2-x^2}{\pi^2+x^2}\tag{1}$$
over the $[\pi,2\pi]$ interval, since the RHS is a decreasing function and its value at $x=2\pi$ is $-\frac{3}{5}<-\frac{1}{2\pi}$. Translating and rearranging, we just need to prove that:
$$ \forall x\in[0,\pi],\qquad \frac{\sin x}{x}\leq \frac{(\pi+x)(\pi+2x)}{\pi^2+(\pi+x)^2}\tag{2}$$
but that is trivial since:
$$ \forall x\in[0,\pi],\qquad \frac{\sin x}{x}\leq \color{red}{1} \leq \frac{(\pi+x)(\pi+2x)}{\pi^2+(\pi+x)^2}.\tag{3}$$
A: Hint: You can make $f(x)=(\pi^2+x^2)\sin x-(\pi^2-x^2)x$. So $f'(x)=2x\sin x+(\pi^2+x^2)\cos x+3x^2-\pi^2$. See that $f'(\pi)=0$, calculate $f''$ and show that $f''(\pi)>0$. Next try to see if $\pi$ is a minimum of f. Since $f(\pi)=0$ the results follows.
A: Let f(x)=sinx-(((π²x-x³))/(π²+x²)),x≥π. Then f(π)=0,f(2π) is positive. Now if f(x)<0 for some x>π then by Rolle's theorem f'(y)=0 in the domain. But then 
0= f ′(y)=cos y-1+((2π²(π²-y²))/((y²+π²)²)) implies that cos y=1-((2π²(π²-y²))/((y²+π²)²))>1, a contradiction.

