<s>[INST] Prove $\frac{x}{\sqrt{1+x^2}} \lt \arctan x$ for every $x \gt 0$ 
Prove $$\frac{x}{\sqrt{1+x^2}} \lt \arctan x$$ for every $x \gt 0$.

I proved half of it with lagrange rule but that I got stuck. Any ideas?
I can upload my work if you want.
 [/INST] let be 
$\displaystyle f:[0,\infty) \rightarrow \mathbb{R} $
$f(x)=\frac{x}{\sqrt{1+x^2}} - \arctan x$
$\displaystyle f'(x)=\frac{x' \cdot \sqrt{1+x^2} - x \cdot (\sqrt{1+x^2})'}{(\sqrt{1+x^2})^2} -(\arctan x)'$
$\displaystyle f'(x)=\frac{1 \cdot \sqrt{1+x^2} - x \cdot \frac{1}{2 \sqrt{1+x^2}}\cdot 2 \cdot x}{(\sqrt{1+x^2})^2} - \frac{1}{1+x^2}$
$\displaystyle f'(x)=\frac{ \sqrt{1+x^2} - \frac{x^2}{\sqrt{1+x^2}} }{{1+x^2}} - \frac{1}{1+x^2}$
$\displaystyle f'(x)=\frac{ \frac{1+x^2}{\sqrt{1+x^2}} - \frac{x^2}{\sqrt{1+x^2}} -1}{{1+x^2}} $
$\displaystyle f'(x)=\frac{ 1+x^2-x^2 -\sqrt{1+x^2} }{{(1+x^2) \cdot \sqrt{1+x^2}}} $
$f'(x)=\frac{1-\sqrt{1+x^2}}{(1+x^2)\sqrt{1+x^2}} <0$
because $f'(x) <0$ we have that $f$ is decreasing on $ [0,\infty)$ 
and thus
$f(0)>f(x)$ for any $x$ in $ (0,\infty)$
but $f(0)=0$, so
$0>f(x)$
$0>\frac{x}{\sqrt{1+x^2}} - \arctan x$
$\arctan x>\frac{x}{\sqrt{1+x^2}}$
</s>