TITLE: Using mean value theorem to prove $x/(1+x^2)<\arctan x<x$
QUESTION [6 upvotes]: Prove using mean-value theorem that $x/(1+x^2)<\arctan x<x$ for $x>0$   
I got the first part but how do I prove $\arctan x< x$  using the MVT?   
The first part was done easily by applying MVT on $\arctan x$, should I use $\arctan x-x$ for the second part? Thanks!

REPLY [8 votes]: I think you have it. One application of the MVT gives you both parts:
Let $x>0$. Applying the Mean Value Theorem to $f(x)=\arctan x$ on the interval $[0,x]$ gives a number $c$ with $0<c<x$ such that
$$
{\arctan x-\arctan 0\over x-0}={1\over 1+c^2}
$$
Rearranging the above gives:
$$
\arctan x={x\over 1+c^2} .
$$
for some $c$ between $0$ and $x$.
Since  $x>c$ and $x\gt0$, we have: $${x\over 1+x^2}\lt{x\over 1+c^2}<x;$$
whence
$$
 {x\over 1+x^2}\lt\arctan x\lt x .
$$