<s>[INST] $x-\sin(x) \geq \dfrac{x^3}{(x+\pi)^2}$ Let $x \geq 0.$ I need to prove that $x-\sin(x)\geq\dfrac{x^3}{(\pi+x)^2}.$
I tried the derivative, of $f(x)=x-\sin(x)-\dfrac{x^3}{(\pi+x)^2}$ which is $1-\cos(x)-\dfrac{x^2(x+3\pi)}{(\pi+x)^3},$ but it has a complicated formula. 
Any ideas, hints?
Edit: sorry, there was a mistake in the derivative, I corrected it.
 [/INST] Put $x=z-\pi$; it suffices to show the inequality for for $z\geq \pi$.
$$
z-\pi -\sin(z-\pi)\geq \frac{(z-\pi)^3}{z^2}
$$
$$
z-\pi +\sin(z)\geq z -3\pi+\frac{3\pi^2}{z} -\frac{\pi^3}{z^2}
$$
$$
2\pi +\sin(z)\geq \frac{3\pi^2}{z} -\frac{\pi^3}{z^2}
$$Note that $\sin(z)\geq -1$. If we set
$$
2\pi- 1=\frac{3\pi^2}{z} -\frac{\pi^3}{z^2},
$$the last solution is $z=\frac{3 π^2 + π^{3/2} \sqrt{4 + π}}{4 π - 2}(:=z_0)\approx 4.21$. But on $(\pi,z_0)$, we can use the approximation $\sin(z)> (\pi-z)-\frac{(\pi-z)^3}{7}$, whence
$$
2\pi +\sin(z)  - \frac{3\pi^2}{z} +\frac{\pi^3}{z^2}
$$
$$
> 2\pi + (\pi-z)-\frac{(\pi-z)^3}{7}-\frac{3\pi^2}{z} +\frac{\pi^3}{z^2}
$$
$$
=\frac{(z-\pi )^3 \left(z^2-7\right)}{7 z^2}>0
$$
So there are no solutions on $(\pi,z_0)$, we don't actually have equality at $z_0$, and we know there are no solutions for $z>z_0$.
</s>