Find the minimum values of $a^3$ and $b^3$ from the following given cubic function.

Question

Today after teaching AM,GM,HM relations my teacher proposed this question.
It isn't a HW though, since we don't have to tell that we solved the question and submit a solution.

Question

Given that $a,b$ are two positive real numbers.
If $$f(x)=2x^3+ax^2+bx+4=0$$
Find the minimum values of $a^3$ and $b^3$.

Well I have progressed a bit.
Applying $AM\ge GM${taking all the terms in $f(x)$ positive}
We get $8x^6ab\le0$ which is impossible since $x$ can't be $0$ and $a,b$ are positive.
So we have a conclusion that roots of $f(x)$ are negative.
Now putting $x=-X$ where $X$ is positive real number, we get
$2X^3+bX=aX^2+4$

This is all useful I got. Using AM,GM on the both the sides one by one yields two inequalities of no use.
Thanks for any solution or hint in advance.

Answer 1

Here's a sketch.

Assume $a, b \geq 0$. Let the roots be denoted $x_1, x_2, x_3$.

Using Vieta's formula we find \begin{align} x_1+x_2+x_3 &= -\frac{a}{2} \\ x_1 x_2 + x_2 x_3 +x_3 x_1&= \frac{b}{2} \\ x1 x_2 x_3 &= -2 \end{align} The last inequality implies that $$2 = (-x_1)(-x_2)(-x_3) = |x_1||x_2||x_3|$$

Applying AM-GM, then \begin{align} \frac{a}{6} &= -\frac{1}{3}(x_1+x_2+x_3) \\ &\geq ((-x_1)(-x_2)(-x_3))^{1/3} \\ &= \ldots \end{align}

Also, \begin{align} \frac{b}{6} &= \frac{1}{3}(x_1 x_2 + x_2 x_3 +x_3 x_1)\\ &\geq (x_1 x_2 x_1 x_3 x_2 x_3)^{1/3}\\ &\vdots \end{align}

Then, once you have found $a, b$ above the minimums will follow.

Answer 2

The minimum value of a positive number does not exist, which says that a minimal value of $a^3$ does not exist and a minimum value of $b^3$ does not exist.