Find Min of function $x+\frac{2}{x}$ for x>0 with following solution?

Question

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.

Answer 1

Since $x+\frac2x>0$, we can assume that $x+\frac2x$ has a non-negative lower bound $c$ for $x>0$, that is: $$\exists c\ge 0\;\forall x>0\;x+\frac2x\ge c$$

This inequality is equivalent to $$\exists c\ge 0\;\forall x>0\;x^2-cx+2\ge 0$$

The polynomial $x^2-cx+2$ has no negative root, because the sum of the roots is $c(\ge 0)$ and their product is $2$.

So, under the assumption $x^2-cx+2\ge0\;\forall x>0$, the polynomial has one root or no root. As you has pointed, that fact is equivalent to this: $$c^2-8\le 0$$ or (remember that $c\ge 0$) $$0\le c\le 2\sqrt 2$$

To sum up:

The real number $c$ is a lower bound for $\{x+\frac2x:x>0\}$ if and only if $0\le c\le 2\sqrt 2$.

This means that $c$ is the greatest lower bound for $x+\frac 2x$. Since $$\sqrt 2+\frac2{\sqrt 2}=2\sqrt 2$$ this lower bound is reached, and thus, it is the minimum that you are looking for.

Answer 2

Using $AM-GM$ inequality, we have(since $x>0$)

$$ x + \frac{2}{x} \geq 2 \sqrt{ x \cdot \frac{2}{x} } = 2 \sqrt{2}$$

Answer 3

Since $x\ge 0$ $$ax+\frac{b}{x} \ge c $$ $$ax^2-cx+b \ge 0$$ Since the quadratic is always greater than equal to zero, so the discriminant $D \le 0$ $$c^2-4ab\le 0$$ As we can we observe that function is decreasing in $(0,\sqrt {2})$ and increasing in $[\sqrt {2},\infty)$ this show that the point of minima must exists at $ \sqrt {2}$

Answer 4

Let's look at the original inequality $$x + \frac{2}{x} \ge c$$ after you have concluded that $$-2\sqrt{2} \le c \le 2\sqrt{2}.$$ If we suppose $c = 0$, then is the original inequality true? Yes: $x + 2/x \ge 0$ is a true statement for $x > 0$. Indeed, any value of $c$ in the interval you obtained, when substituted into the first inequality, results in a true statement, by construction. But all of the inequalities resulting from choosing $c < 2 \sqrt{2}$ are looser than the inequality resulting from choosing $c = 2 \sqrt{2}$, in the sense that for any $c' < 2 \sqrt{2}$, we trivially have $$x + \frac{2}{x} \ge 2 \sqrt{2} > c',$$ hence $c = 2 \sqrt{2}$ results in the tightest inequality of the set of choices we can make.

The only missing part is to show that equality can be attained for this choice of $c$ thus guaranteeing that the inequality is strict, and of course, this occurs when $x = 2/x = \sqrt{2}.$