If $ax^2+\frac{b}{x}\ge c$ for all positive $x$ where $a>0$ and $b>0,$ then show that $27ab^2\ge4c^3$ If $ax^2+\frac{b}{x}\ge c$ for all positive $x$ where $a>0$ and $b>0,$ then show that $27ab^2\ge4c^3$

I considered the function $f(x)=ax^2+\frac{b}{x}$ and I put $f'(x)=0$ to find $x^3=\frac{b}{2a}$
I am stuck here.

By AM-GM $$ax^2+\frac{b}{x}=ax^2+2\left(\frac{b}{2x}\right)\geq3\sqrt[3]{ax^2\cdot\left(\frac{b}{2x}\right)^2}=3\sqrt[3]{\frac{ab^2}{4}}.$$
Thus, we need $$3\sqrt[3]{\frac{ab^2}{4}}\geq c,$$
which gives which you wish.
