<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] Maybe an easier way to do it, though conceptually similar with @Integrand
By multiplying out and simplifying the terms, while noting that $x(x+\pi)^2-x^3=\pi x(x+2\pi)=\pi(x+\pi)^2-\pi^3$, the inequality is equivalent to:
$(\pi-\sin x)(x+\pi)^2 \ge \pi^3, x \ge 0$
But now $\pi-\sin x \ge \pi -1$ and we see that $(\pi+1)^2(\pi-1)=(\pi^2-1)(\pi+1)=\pi^3+\pi^2-\pi-1>\pi^3$, so the inequality is automatically true for $x \ge 1$
Then taking derivatives for $f(x)=(\pi-\sin x)(x+\pi)^2$ we have $f'(x)=2(x+\pi)(\pi-\sin x)-(x+\pi)^2 \cos x$ and for $0 \le x \le 1$, $2(\pi-\sin x)-(x+\pi) \cos x \ge 2\pi-2 -\pi-1= \pi -3 >0$ so $f$ is increasing there, hence $f(x) \ge f(0)=\pi^3$ and the inequality holds in this interval too. so done!
</s>