Proving that $\sin x \ge \frac{x}{x+1}$

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

$$ \sin(x) \geq \frac{x}{\sqrt{x^2 + 1}} \geq \frac{x}{x + 1}. $$

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

$$ f(x)=(x+1)\sin x-x\geq 0=f(0)$$

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 $$ 1-\frac{1}{6}x^2-\frac{1}{x+1}=\frac{1}{6}\cdot\frac{x(6-x^2-x)}{x+1}\geq 0. $$