prove：$\left|x+\frac{1}{x}\right|\geq 2$ prove：
$\left|x+\frac{1}{x}\right|\geq 2$
Can I just use 
$\left|\left(\sqrt{x}\right)^2+\left(\frac{1}{\sqrt{x}}\right)^2\right|\geq 2$
and
$\left|\left(\sqrt{-x}\right)^2+\left(\frac{1}{\sqrt{-x}}\right)^2\right|\geq 2$?

Denote $x+\frac{1}{x}=v$. 
Obviously $|v|>2$ if either $v>2$ or $-v<-2$. The first case is therefore $x +\frac{1}{x}>2$ and it becomes $(x-1)^2>0$, which is always true. The second case is $-x-\frac{1}{x}<-2$ or $\frac{x^2-2x+1}{x}>0$ and also breaks down into 2 cases: either $(x-1)^2>0 \cap x>0$ or $(x-1)^2<0 \cap x<0$. Obviously the second case is never true and the first case is always true. 
Hence $|x+\frac{1}{x}|$ is always larger than $2$.
