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.

We have $x=(\frac {b}{2a})^{\frac {1}{3}} $.  From second derivative we find that at this point we have a minima as $f''(x)>0$ . Thus rearranging original equation we have $ax^3+b\geq cx $ plugging in value we have $\frac {3b}{2}\geq c.(\frac {b}{2a})^{\frac {1}{3}} $ cubing and rearranging $27ab^2\geq 4c^3$. Hence the proof.
