$$y=(\sqrt{x}+\frac{1}{\sqrt{x}})^{2}=$$
$x=\frac{-\sqrt{y\left(y-4\right)}+y-2}{2}$
$x=\frac{\sqrt{y\left(y-4\right)}+y-2}{2}\text{, }y\geq 4$
$y=\frac{\left(x+1\right)^{2}}{x}$
$x>0$
