Prove: for all $x\neq 0,$ $\left\lvert x+\frac{1}{x}\right\rvert > 2$ and $\left\lvert x+\frac{1}{x}\right\rvert = 2\iff x=\pm 1$
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 $x\neq 1$
Case 2: $x< 0$
In the same manner we get $$x^2+2x+1>0$$
and $x\neq -1$
Now we need that both cases will occur simultaneously (?) so we get $x=\neq \pm 1$ for an equality and $x\neq 0$ for inequality
It's just AM-GM: $$\frac{x^2+1}{2}\geq|x|,$$ which for $x\neq0$ gives which you wish: $$\frac{x^2+1}{|x|}\geq2$$ or $$\left|x+\frac{1}{x}\right|\geq2.$$