Find Min of function $x+\frac{2}{x}$ for x>0 with following solution? I want to find Minimum of function $x+\frac{2}{x}$ for $x>0$ with following solution. My problem is how can we say we found the answer with following solution.
Solution:
Let Minimum be something like c. We have:
$x+\frac{2}{x} \ge c \to \frac{x^2+2}{x} \ge c \to x^2 + 2 \ge cx \to x^2 -cx +2 \ge 0 $
So $\Delta \le 0$ because the quadratic must not have two solutions so it will be only positive or zero. So:
$\Delta = c^2 - 8 \le 0 \to c^2 \le 8 \to -2\sqrt{2} \le c \le +2\sqrt{2}$
My question is how can we say with following solution that the minimum  of $x+\frac{x}{2}$ must be $+2\sqrt{2}$? We find the maximum of $c$, How can we say that maximum is minimum of function? Is it a valid solution?
Thanks and sorry for my English.

Using $AM-GM$ inequality, we have(since $x>0$)
$$ x + \frac{2}{x} \geq 2 \sqrt{ x \cdot \frac{2}{x} } = 2 \sqrt{2}$$
