Cannot understand how does he find the minimum value of this funtion My teacher has find the minimum value of $y = x + \frac{2}{x}$ when $x \ge 0$ without using 
differential:
$y = x + \frac{2}{x}$
Multiplied both side by $x$ and moved $y$ to other side
$x^2 - yx + 2 = 0$
Then he wrote that having an answer in this equation is conditioned by
$y^2 - 8 \ge 0$  (He used $b^2 - 4ac$ formula)
$y \ge 2\sqrt{2}$
The answer is $2\sqrt{2}$ and it's correct.
Can anyone explain this solution?

If the discriminant is smaller than zero($b^2-4ac<0$) than $x_{1,2}\in\mathbb{C}$ so clearly $b^2-4ac\geq0$ now if $y\in(-2\sqrt{2},2\sqrt{2})$ it means that there is no solution in reals,and since $x>0$ than $y>0$ so we have that $y\geq2\sqrt{2}$ clearly since we are looking for the minimum we pick the smallest $y$ which is $y=2\sqrt{2}$ And since we have a positive $x$ that satisfies that equation we are done.
