# Simple absolute value inequality Prove: for all x!=0, ∣∣x+1/x∣∣>2 and ∣∣x+1/x∣∣=2<=>x=+-1 case 1: x>0 x+1/x>2 x^2−2x+1>0 x_(1,2)=((2+sqrt(4−4))/2=1 Therefore x!=1 Case 2: x<0 In the same manner we get x^2+2x+1>0 and x!=−1 Now we need that both cases will occur simultaneously (?) so we get x!=+-1 for an equality and x!=0 for inequality

Simple absolute value inequality
Prove: for all $x\ne 0,$$|x+\frac{1}{x}|>2$ and $|x+\frac{1}{x}|=2\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}x=±1$
case 1: $x>0$
$x+\frac{1}{x}>2$
${x}^{2}-2x+1>0$
${x}_{1,2}=\frac{2+\sqrt{4-4}}{2}=1$
Therefore $x\ne 1$
Case 2: $x<0$
In the same manner we get
${x}^{2}+2x+1>0$
and $x\ne -1$
Now we need that both cases will occur simultaneously (?) so we get $x=\ne ±1$ for an equality and $x\ne 0$ for inequality
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Jayleen Copeland
It's just AM-GM:
$\frac{{x}^{2}+1}{2}\ge |x|,$
which for $x\ne 0$ gives which you wish:
$\frac{{x}^{2}+1}{|x|}\ge 2$
or
$|x+\frac{1}{x}|\ge 2.$