Proving that $\sin x \ge \frac{x}{x+1}$ Prove that
$$ \sin x \ge \frac{x}{x+1}, \space \space\forall x \in \left[0, \frac{\pi}{2}\right]$$
 A: 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
$$ f(x)=(x+1)\sin x-x\geq 0=f(0)$$
A: 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
$$
1-\frac{1}{6}x^2-\frac{1}{x+1}=\frac{1}{6}\cdot\frac{x(6-x^2-x)}{x+1}\geq 0.
$$
A: 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
$$
\sin(x) \geq \frac{x}{\sqrt{x^2 + 1}} \geq \frac{x}{x + 1}.
$$ 
A: Inequality is easily verified for $x=0,1$
$\csc x - \cot x = \frac{1}{\csc x + \cot x}<1 $ for $x \in \left(0, \frac{\pi}{2} \right)$
Hence, $\csc x < 1 +\cot x < 1+ \frac{1}{x} $ since its well known that $\tan x>x$
and this leads to the desired inequality
