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

Question

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

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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.

Answer 1

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.

Answer 2

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.