Rational bounds for $\tan^{-1} x$ I want help with this question.

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

Thank you.
 A: Let $t:=\tan^{-1}x$ so that $x=\tan(t)$. Then $1+x^2=\displaystyle\frac{\cos^2(t)+\sin^2(t)}{\cos^2(t)}=\frac1{\cos^2(t)}$, so
$$\frac x{1+x^2}=\tan(t)\cdot\cos^2(t)=\sin(t)\cos(t)=\frac{\sin(2t)}2\,.$$
So this leads to prove $\displaystyle\frac{\sin(2t)}2<t<\tan(t)$ for $t\in (0,\pi/2)$. 
A: 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. :-)
A: $$ \frac{d}{dx} \tan^{-1} x  = \frac{1}{1+x^2} < \frac{d}{dx} x =1$$
and $\tan^{-1}(0)=0$ so that the last inequality is proved. 
$$  \frac{d}{dx} \frac{x}{x^2+1} = \frac{(x^2+1) - x(2x) }{(x^2+1)^2} = \frac{1-x^2}{(x^2+1)^2} < \frac{d}{dx} \tan^{-1} x  = \frac{1}{1+x^2} $$
and $\frac{x}{1+x^2}(0)=0=\tan^{-1}(0)$ so that the first inequality is proved. 
Another way : 
$$\frac{x}{x^2+1}< t\frac{1}{t^2+1}|_{t=0}^{t=x} - 
\int_0^x t\frac{-2t}{(t^2+1)^2} = \int_0^x \frac{1}{t^2+1}\ dt =\tan^{-1}x < \int_0^x \frac{1}{0+1}\ dt=x$$
