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.

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
$$\arctan(x)\ge \frac{x}{\sqrt{1+x^2}}$$
