How to prove $\frac{x}{1+x^2}<\arctan x<x$ for $x>0$?

We can use the Mean Value Theorem: $$ \frac{\tan^{-1}(x)-\tan^{-1}(0)}{x-0}=\frac1{1+\xi^2} $$ for some $\xi\in(0,x)$. That is $$ \frac1{1+x^2}\lt\frac{\tan^{-1}(x)}{x}\lt1 $$

For example:

$$f(x):=\frac x{1+x^2}-\arctan x\implies f'(x)=\frac{1-x^2}{(1+x^2)^2}-\frac1{1+x^2}=-\frac{2x^2}{(1+x^2)^2} $$

Since clearly $\;f'(x)\le 0\;\;\forall\,x\ge0\;$ , and in fact $\;f'(x)<0\;\;\forall\,x>0\;$ , the function is monotone descending, so

$$\forall\;x>0\;,\;f(x)<f(0)=0\;\;\text{and we're done with the leftmost inequality}$$

Now you try the rightmost inequality (very similar trick)

\begin{align} & \frac x {1+x^2} = \int_0^x \frac 1 {1+x^2} \, dt \le \int_0^x \frac 1 {1+t^2} \,dt \\[8pt] = {} & \arctan x = \int_0^x \frac{dt}{1+t^2} \le \int_0^x 1\,dt = x. \end{align}