hazbijav6

2022-09-05

Simple absolute value inequality

Prove: for all $x\ne 0,$$|x+\frac{1}{x}|>2$ and $|x+\frac{1}{x}|=2\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}x=\pm 1$

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 $x=\ne \pm 1$ for an equality and $x\ne 0$ for inequality

Prove: for all $x\ne 0,$$|x+\frac{1}{x}|>2$ and $|x+\frac{1}{x}|=2\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}x=\pm 1$

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 $x=\ne \pm 1$ for an equality and $x\ne 0$ for inequality

Jayleen Copeland

Beginner2022-09-06Added 4 answers

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

$|x+\frac{1}{x}|\ge 2.$

$\frac{{x}^{2}+1}{2}\ge |x|,$

which for $x\ne 0$ gives which you wish:

$\frac{{x}^{2}+1}{|x|}\ge 2$

or

$|x+\frac{1}{x}|\ge 2.$

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