Is it true that $\sin x > \frac x{\sqrt {x^2+1}} , \forall x \in (0, \frac {\pi}2)$? Is it true that $$\sin x > \dfrac x{\sqrt {x^2+1}} , \forall x \in \left(0, \dfrac {\pi}2\right)$$  (I tried differentiating , but it's not coming , please help) 

Square both sides (assume $x\in(0,\pi/2)$):
$$\sin x>\frac{x}{\sqrt{x^2+1}}\iff 1-\cos^2 x>1-\frac{1}{x^2+1}$$
$$\iff \cos ^2 x<\frac{1}{x^2+1}=\frac{\sin^2 x+\cos^2 x}{x^2+1}$$
$$\iff x^2\cos^2 x+\cos^2 x<\sin^2 x + \cos^2 x$$
$$\iff \tan^2 x> x^2\iff \tan x>x,$$
which is true ($(\tan x-x)'>0$ for $x\in(0,\pi/2)$ and $\tan(0)-0=0$).
