$x-\sin(x) \geq \dfrac{x^3}{(x+\pi)^2}$ 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.

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 $x\gt0$,
$$
x-\sin(x)\ge\frac{x^3}{(x+\pi)^2}\tag9
$$
