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) 
 A: How about the substitution $x = \tan \theta$, then the inequality reduces to proving:
$$\sin \tan \theta \ge \sin \theta$$
for, $\theta \in \left(0,\tan^{-1} (\pi/2\right))$.
Since, $\sin \theta$ is monotone increasing in the interval $\left(0,\pi/2\right)$, we are done using: $$\tan \theta \ge \theta$$
A: 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 $\tan(0)-0=0$).
A: $\arcsin$ is an increasing function, and you can rewrite
$$x>\arcsin\left(\frac{x}{\sqrt{x^2+1}}\right).$$
Then, deriving (and using equality at $x=0$)
$$1>\frac{\frac{\sqrt{x^2+1}-\frac{x^2}{\sqrt{x^2+1}}}{x^2+1}}{\sqrt{1-\left(\frac{x}{\sqrt{x^2+1}}\right)^2}}=\frac1{x^2+1}.$$

The original inequation is a mixture of trigonometric and algebraic expressions, which makes it intractable as such. The "trick" is to let the sine disappear, knowing that the derivative of the $\arcsin$ is an algebraic expression.
