# Using mean value theorem to prove $x/(1+x^2)<\arctan x<x$ [closed]

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!

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## closed as off-topic by 6005, Claude Leibovici, USER91500, Alex M., Tom-TomNov 30 '15 at 13:53

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– Martin Sleziak Nov 30 '15 at 9:15

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 $${x\over 1+x^2}\lt\arctan x\lt x .$$