Answer
$x + \frac{1}{x} \ge 2$ (and $x\gt 0$)
Work Step by Step
We know that $x^2 \ge 0$ for any real number $x$.
Whenever $x>0$, then we can write:
$(\sqrt{x} - \frac{1}{\sqrt{x}})^2 \ge 0$
Thus
$x+\frac{1}{x} -2 \sqrt{x} \times \frac{1}{\sqrt{x}} = x + \frac{1}{x} -2 \ge 0$.
Hence,
$x + \frac{1}{x} \ge 2$ (and $x\gt 0$)
