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

I proved earlier (see my answer for $\lim_{x \to 0}{\frac{x-\sin{x}}{x^3}}=\frac{1}{6}$)
that
$$
\sin\,x\geq x-\frac{1}{6}x^3,\quad x\in[0,\pi/2].
$$
From this
$$
\frac{\sin\,x}{x}\geq 1-\frac{1}{6}x^2.
$$
Now for $x\in[0,\pi/2]$ we have
$$
1-\frac{1}{6}x^2-\frac{1}{x+1}=\frac{1}{6}\cdot\frac{x(6-x^2-x)}{x+1}\geq 0.
$$
