Show that $\frac{ -1 }{ 2 } \le \frac{ x }{ 1+x^2 } \le \frac{ 1 }{ 2 }$ So I'm trying to show that:
$\dfrac{ -1 }{ 2 } \le \dfrac{ x }{ 1+x^2 } \le \dfrac{ 1 }{ 2 }$
for every value of x.
I know I have to use mean value theorem so I tried to show it with cases. First I tried showing that $\dfrac{-1}{2} \le \dfrac{x}{1+x^2}$ and then $\dfrac{1}{2} \ge \dfrac{x}{1+x^2}$ using MVT. Is that correct? Can someone guide me?
Thanks!

$$0\leq(1-x^2)^2\iff 4x^2\leq(1+x^2)^2\iff\frac{x^2}{(1+x^2)^2}\leq\frac14
\iff\left|\frac{x}{1+x^2}\right|\leq\frac12.$$
