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.
 A: $$x^2+\frac{1}{x^2}\ge 2\sqrt{x^2\cdot \frac{1}{x^2}}=2\Rightarrow x^2+2+\frac{1}{x^2}\ge 4\Rightarrow \left(x+\frac 1x\right)^2\ge 2^2\Rightarrow \left|x+\frac 1x\right|\ge 2.$$
A: Note that $|x+{1 \over x}| = |(-x)+{1 \over (-x)}|$, hence we need only consider $x>0$.
The function $f(x) = x+{1 \over x}$ is strictly convex on $(0,\infty)$ and $f'(1) = 0$, hence $1$ is a minimiser. Hence $f(x) \ge f(1)$ for all $x >0$. Since $f(1) = 2$, you have your answer.
A: Since both sides are nonnegative, it suffices to prove the squared version:
$$
(x + \tfrac{1}{x})^2 \geq 4
$$
Indeed, observe that:
\begin{align*}
(x + \tfrac{1}{x})^2
&= x^2 + 2 + \tfrac{1}{x^2} \\
&= (x^2 - 2 + \tfrac{1}{x^2}) + 4 \\
&= (x - \tfrac{1}{x})^2 + 4 \\
&\geq 0 + 4 &\text{since } x - \tfrac{1}{x} \in \mathbb R\\
&= 4
\end{align*}
as desired.
A: 
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 
$$\left|x + \frac1{x}\right| = -x - \frac1{x} = \left( -x - \frac1{x} - 2 \right) + 2 = \frac{(x+1)^2}{-x}+2 \ge 2.$$

