Is the theory of $\Delta\le0$, true in cubic functions$?$

Question

If $ax^2+\frac{b}{x}\ge c$ $\forall x>0$ where $a>0 \:\: , b>0$ Show that $27ab^2\ge4c^3$

My work:

Let a function $f(x)$ be $$ax^2+\frac{b}{x}\ge c$$ or we can rewrite $f(x)$ as $$ax^3-cx+b\ge0$$ and also $f(x)=$ $$x^3-\frac{c}{a}\,x+\frac{b}{a}\ge0$$ Now since the cubic function $f(x)$ is always $\ge0$ therefore $\Delta$ should be $\le0$ then I am able to prove the desired result. But why is it true$?$ Because the theory of

$\Delta \le0$ if $f(x)\ge0$ (assuming that the leading coefficient is positive) is only applicable when $f(x)$ is a quadratic polynomial, in my opinion.

Or is it applicable in any polynomial of any degree$?$ Or is there any other, better way to do it$?$

Any help is greatly appreciated.

Answer 1

Since all numbers are positive, by the AM-GM inequality:

$$\require{cancel} ax^2+\frac{b}{x} = ax^2 + \frac{b}{2x} + \frac{b}{2x} \;\ge\; 3 \sqrt[3]{a\cancel{x^2} \cdot \frac{b}{2 \cancel{x}} \cdot \frac{b}{2 \cancel{x}}} = 3 \sqrt[3]{\frac{ab^2}{4}} \tag{1}$$

For the $ax^2+\frac{b}{x}\ge c$ inequality to hold for $\forall x>0$ the minimum on the RHS of $(1)$ must be $\ge c\,$:

$$ 3 \sqrt[3]{\frac{ab^2}{4}} \ge c \;\;\iff\;\; 27ab^2 \ge 4 c^3 $$

Answer 2

Instead of looking at the discriminant of the cubic, let's use a different approach.

We search for the extremal value of the function $$f(x)=ax^2+\frac{b}{x}.$$ Using differentiation this is found at $$0=f'(x_0)=2ax_0-\frac{b}{x_0^2}\iff x_0=\sqrt[3]{\frac{b}{2a}}$$ thus it equals $$f\left(x_0\right)=\frac{ax_0^3+b}{x_0}=\frac{a\cdot \frac{b}{2a}+b}{\sqrt[3]{\frac{b}{2a}}}=\frac{3}{2}\frac{b}{\sqrt[3]{\frac{b}{2a}}}.$$

Now the assumtion states that this must be greater or equal than $c$, so $$\frac{3}{2}\frac{b}{\sqrt[3]{\frac{b}{2a}}}\geq c$$ and rearranging terms gives the looked for inequality.