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!
 A: WLOG $x=\tan y$
$$\implies\dfrac{2x}{1+x^2}=\sin2y$$
Alternatively,
$$\dfrac x{1+x^2}=\dfrac1{x+\dfrac1x}$$
Now if $x>0,\dfrac{x+\dfrac1x}2\ge\sqrt{x\cdot\dfrac1x}=1$ 
What if $x<0?$
A: I'm not sure why mean value theorem is required as it is much easier to do it like this - 
For any $x\in\mathbb{R}$ we know that $$(1+x)^2\geq0\Longrightarrow 1+x^2+2x\geq0\Longrightarrow 2x\geq -(1+x^2)\Longrightarrow \dfrac{x}{1+x^2}\geq-\dfrac{1}{2}$$ Similarly $$(1-x)^2\geq0\Longrightarrow 1+x^2-2x\geq0\Longrightarrow1+x^2\geq2x\Longrightarrow\dfrac{1}{2}\geq\dfrac{x}{1+x^2}$$
And we're done. The only thing we used is that the square of a real number is positive.
A: Let $u=|x|$, then $\frac{ -1 }{ 2 } \le \frac{ x }{ 1+x^2 } \le \frac{ 1 }{ 2 } \iff  \frac{ u }{ 1+u^2 } \le \frac{ 1 }{ 2 }$.
You are using $AM-GM$ here and not Mean Value Theorem since $1+u^2\geq2u$ by $AM-GM$.
To do the calculus way, just find the derivative of $x\over {1+x^2}$.
$df\over dx$$={1+x^2-2x^2\over{(1+x^2)^2}}=0\implies x=\pm1$.
Sub in $x=1$ you get $f={1\over2}$ and sub in $x=-1$ you get $f={-{1\over2}}$.
And $lim_{x\to \infty}f=0$ shows it is a global maximum. Minimum similarly.
A: $$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.$$
