$x-\sin(x) \geq \dfrac{x^3}{(x+\pi)^2}$ 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.
 A: 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!
A: The product formula for sine says
$$
\frac{\sin(x)}{x}=\prod_{k=1}^\infty\left(1-\frac{x^2}{k^2\pi^2}\right)\tag1
$$
Thus, for $0\lt x\le\pi$
$$
\begin{align}
\frac{\sin(x)}x
&\le1-\frac{x^2}{\pi^2}\tag2\\[3pt]
&\le1-\frac{x^2}{(x+\pi)^2}\tag3\\
x-\sin(x)
&\ge\frac{x^3}{(x+\pi)^2}\tag4
\end{align}
$$
Explanation:
$(2)$: follows from $(1)$
$(3)$: $x\gt0$
$(4)$: subtract from $1$ (which reverses the inequality)
$\phantom{\text{(4):}}$ and multiply by $x$

For $x\gt\pi$,
$$
\begin{align}
(x-\sin(x))(x+\pi)^2
&\ge(x-1)(x+\pi)^2\tag5\\[3pt]
&=x^3+(2\pi-1)x^2+\left(\pi^2-2\pi\right)\!x-\pi^3\tag6\\[3pt]
&\ge x^3\tag7\\
x-\sin(x)
&\ge\frac{x^3}{(x+\pi)^2}\tag8
\end{align}
$$
Explanation:
$(5)$: $\sin(x)\lt1$
$(6)$: expand product
$(7)$: plug in $x\gt\pi$
$(8)$: divide by $(x+\pi)^2$

Inequalities $(4)$ and $(8)$ together show that for $x\gt0$,
$$
x-\sin(x)\ge\frac{x^3}{(x+\pi)^2}\tag9
$$
A: 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 $z>z_0$.
