Proving that $\sin x \ge \frac{x}{x+1}$
Solution 1:
Take $x \in [0, \pi/2]$. Consider the right triangle with sides $1, x$ and $\sqrt{1 + x^2}$. The angle opposite the side with length $x$ is smaller than $x$. It follows that
$$ \sin(x) \geq \frac{x}{\sqrt{x^2 + 1}} \geq \frac{x}{x + 1}. $$
Solution 2:
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
$$ f(x)=(x+1)\sin x-x\geq 0=f(0)$$