Proving that $\sin x \ge \frac{x}{x+1}$ Prove that
$$ \sin x \ge \frac{x}{x+1}, \space \space\forall x \in \left[0, \frac{\pi}{2}\right]$$

On the given interval:
$$f(x):=(x+1)\sin x-x\Longrightarrow f'(x)=\sin x+(x+1)\cos x-1\geq 0$$
since $\,\sin x+(x+1)\cos x\geq 1$ on  $\,[0,\pi/2]\,$.
Thus, $\,f\,$ is monotone non-descending on $\,\left[0,\dfrac{\pi}{2}\right]\,$ and thus
$$ f(x)=(x+1)\sin x-x\geq 0=f(0)$$
