Prove $\frac{x}{\sqrt{1+x^2}}<\arctan\left(x\right)$ for every $x>0$ 
Prove $\frac{x}{\sqrt{1+x^2}}<\arctan\left(x\right)$ for every $x>0$ 

I thought about MVT , what i did is : 
let $g(x) = \arctan\left(x\right) $
so by MVT there is $c\in (0,x)$ such that : 
$g'(c) = \frac{g\left(x\right)-g\left(0\right)}{x-0} \implies \frac{1}{c^2+1} = \frac{\arctan\left(x\right)}{x}$ 
now i need to prove that $\frac{x}{c^2+1}>\frac{x}{\sqrt{x^2+1}}$
any hint how to prove this  ? 
thanks 

Let $f(x)= \arctan x - \frac{x}{\sqrt{1+x^2}}$. Evaluate 
$$f'(x)=\frac{\sqrt{1+x^2}-1}{(1+x^2)^{3/2}}>0$$
So, $f(x)$ is strictly increasing for $x>0$. Thus, $f(x) > f(0) = 0$, or
$$\arctan x > \frac{x}{\sqrt{1+x^2}}$$
