Prove that for $x>0$ we have $\sin{x}<\frac{x}{\sqrt{1+\frac{x^2}{3}}}$ Prove the inequality:
$$\sin{x}<\frac{x}{\sqrt{1+\frac{x^2}{3}}}$$
for real $x>0$.
Any hints?
EDIT: Using calculus and derivatives is perfectly allowed. 

This is an incomplete answer.
Perhaps you can do it easily for $x>\sqrt{\frac{3}{2}}$
$1+\frac{x^2}{3}<x^2 \hspace{1cm} for \hspace{0.1cm}x>\sqrt{\frac{3}{2}}$
Then
$(sin^2x+cos^2x)(1+\frac{x^2}{3})<x^2 \hspace{1cm}$
Therefore
$(sin^2x)(1+\frac{x^2}{3})<(sin^2x+cos^2x)(1+\frac{x^2}{3})<x^2 \hspace{1cm} \implies $
$(sinx)\sqrt{(1+\frac{x^2}{3})}<x \implies \sin{x}<\frac{x}{\sqrt{1+\frac{x^2}{3}}}$
