TITLE: Rational bounds for $\tan^{-1} x$
QUESTION [5 upvotes]: I want help with this question.

Show that for all $x>0$, $$ \frac{x}{1+x^2}<\tan^{-1}x<x.$$

Thank you.

REPLY [2 votes]: I know a method which is based of using The mean value theorem for functions in Calculus. Let $f(x)=\arctan(x)$. Since $f'=1/1+x^2$ so according to mean value theorem, there is a $\xi\in(0,x)$ such that $$\frac{\tan^{-1}(x)-\tan^{-1}(0)}{x-0}=f'(\xi)$$ But $0<\xi<x$ makes $f'(\xi)$ to be: $$f'(x)<f'(\xi)<1$$ I think the rest is clear. Indeed multiplying both sides by $x\neq 0$ gives the result. :-)