TITLE: How to prove $\frac{x}{1+x^2}<\arctan x<x$ for $x>0$?
QUESTION [11 upvotes]: How to prove for $x>0$, $\dfrac{x}{1+x^2}<\arctan x<x$?

I started with saying $0<\arctan x<\frac{\pi}{2}$  when $x>0$
I'm just not sure how to proceed with this proof, it works when I take random  values but I cant see how to prove that it works for any $x$.

REPLY [9 votes]: \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}