TITLE: Is it true that $\sin x > \frac x{\sqrt {x^2+1}} , \forall x \in (0, \frac {\pi}2)$?
QUESTION [6 upvotes]: 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)

REPLY [5 votes]: $\arcsin$ 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.