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
$\left|\left(\sqrt{-x}\right)^2+\left(\frac{1}{\sqrt{-x}}\right)^2\right|\geq 2$?

Without loss of genarality we can consider $x\ge 0$(Reason: $x$ and $\frac{1}{x}$ has same sign so if $x<0$ the in place of $x$ we will take $-x(>0)$ and it will not make any difference since $\left| x+\frac{1}{x}\right |=\left| -x+\frac{1}{-x}\right |$)
Then we have,
$\left| x+\frac{1}{x}\right |=x+\frac{1}{x}=(\frac{1}{\sqrt{x}}-\sqrt{x})^2+2\ge 2$
You can use(with its proof which is done by my above arguement) either one(not both ) of the following(if you want to handle only real stuff) according to $x$ is positive or negative,
$\left|\left(\sqrt{x}\right)^2+\left(\frac{1}{\sqrt{x}}\right)^2\right|\geq 2$ or $\left|\left(\sqrt{-x}\right)^2+\left(\frac{1}{\sqrt{-x}}\right)^2\right|\geq 2$
You cant use both since ,if $x$ is positive then $\sqrt{-x}$ is complex.And if $x<0$ then $\sqrt{x}$ is complex.
