<s>[INST] Showing an Absolute Value Inequality Problem Proof I tried solving this question but it does not works for me.
Q.) Show that $\left|x + \frac1{x}\right| \ge 2$ for all $x \ne 0$
There are two ways to do. One is squaring and other is to use absolute value definition.
But I don't understand.
Please help.
 [/INST] 
One is squaring and other is to use absolute value definition. But I
  don't understand.

So here you go:
1) One is squaring:
Since both sides of the inequality is positive, by squaring, it is equivalent to $$\left|x + \frac1{x}\right|^2 \ge 4$$
Notice that 
$$\left|x + \frac1{x}\right|^2 = \left(x + \frac1{x}\right)^2 = x^2 + 2x\frac{1}{x}+\frac{1}{x^2} = x^2 - 2x\frac{1}{x}+\frac{1}{x^2} + 4 = \left(x - \frac1{x}\right)^2 + 4,$$
which is always greater or equal to $4$.
2) Other is to use absolute value definition:
By definition, $|a|$ equals $a$ if $a\ge 0$ and equals $-a$ if $a<0$. Thus:


*

*If $x>0$ then $x + \frac1{x} >0$, thus 
$$\left|x + \frac1{x}\right| = x + \frac1{x} = \left( x + \frac1{x} - 2 \right) + 2 = \frac{(x-1)^2}{x}+2 \ge 2.$$

*If $x<0$ then $x + \frac1{x} <0$, thus 
$$\left|x + \frac1{x}\right| = -x - \frac1{x} = \left( -x - \frac1{x} - 2 \right) + 2 = \frac{(x+1)^2}{-x}+2 \ge 2.$$

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