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.
 A: Let $x=\tan\theta$ with $0\lt\theta\lt\pi/2$.  Since $\sqrt{1+\tan^2\theta}=\sec\theta$ on the left and $\arctan(\tan\theta)=\theta$ on the right, the given inequality is equivalent to the inequality
$$\sin\theta\lt\theta\quad\text{for }0\lt\theta\lt\pi/2$$
A: METHODOLOGY $1$:
The integral definition of the arctangent function is given by
$$\arctan(x)=\int_0^x \frac{1}{1+t^2}\,dt$$
Noting that for $t\in [0,x]$, $x\ge 0$, $1+t^2\le (1+t^2)^{3/2}$, we can assert
$$\arctan(x)\ge \int_0^x \frac{1}{(1+t^2)^{3/2}}\,dt=\frac{x}{\sqrt{1+x^2}}$$
and we are done!

METHODOLOGY $2$:
Recall from basic geometry the inequality for the sine function 
$$\sin y\le y \tag 1$$
for $y\ge 0$.  From $(1)$ it is easy to see that for $0\le y<1$, we have
$$\cos y\ge \sqrt{1-y^2} \tag 2$$
Taking $(1)$ and $(2)$ together reveals that for $0\le y<1$
$$\tan y\le \frac{y}{\sqrt{1-y^2}} \tag 3$$
Now, let $x=\frac{y}{\sqrt{1-y^2}}$.  Note that for $0\le y<1$, $x>0$ and 
$$y=\frac{x}{\sqrt{1+x^2}} \tag 4$$
Using $(4)$ in $(3)$, we obtain
$$\tan\left(\frac{x}{\sqrt{1+x^2}}\right)\le x$$
Taking the inverse function yields the desired inequality
$$\arctan(x)\ge \frac{x}{\sqrt{1+x^2}}$$
A: Suggestion: Take $f(x) = \arctan x - {x\over (1+x^2)^{1/2}}$. What is $f(0)$? What is the sign of $f'(x)$ for $x> 0 $? Does that give a lower bound on $f(x)$ for $x> 0$?
A: 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
$0>f(x)$
$0>\frac{x}{\sqrt{1+x^2}} - \arctan x$
$\arctan x>\frac{x}{\sqrt{1+x^2}}$
A: For $x=0$ we have equality. Then, deriving, for $x>0$ we have
$$\frac{\sqrt{1+x^2}-\dfrac{x^2}{\sqrt{1+x^2}}}{1+x^2}=\frac1{(1+x^2)^{3/2}}<\frac1{1+x^2}$$ and the LHS grows slowlier than the RHS.

The main "trick" is to use differentiation to let the transcendental function disappear.
