<s>[INST] 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!
 [/INST] 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 $x<0?$
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