If $a{x}^2+\frac{b}{x}\ge c$ for all positive x, where $a,b,>0$, then

Consider the function $f\left(x\right)$ given by 

$f\left(x\right)=a{x}^2+\frac{b}{x}-c,x>0$

Clearly, $f\left(x\right)\ge 0$ for all $x>0$.                         $\left[\because a{x}^2+\frac{b}{x}\ge c\text{ (Given) }\right]$

Also, $f^{\mathrm{\prime }}\left(x\right)=2ax-\frac{b}{x^2}$ and $f^{\prime \prime }\left(x\right)=2a+\frac{2b}{x^3}$

For points of local maximum/minimum, we must have 

$f^{\mathrm{\prime }}\left(x\right)=0\Rightarrow 2a{x}^3=b\Rightarrow x={\left(\frac{b}{2a}\right)}^{1/3}$

Now, $f^{\prime \prime }\left\{{\left(\frac{b}{2a}\right)}^{1/3}\right\}=2a+4a=6a>0$

Therefore, $f\left(x\right)$ attains a local minimum at $x={\left(\frac{b}{2a}\right)}^{1/3}$

Now,

$\begin{array}{l}f\left\{{\left(\frac{b}{2a}\right)}^{1/3}\right\}=a{\left(\frac{b}{2a}\right)}^{2/3}+b{\left(\frac{2a}{b}\right)}^{1/3}-c\\ \Rightarrow f\left\{{\left(\frac{b}{2a}\right)}^{1/3}\right\}=\frac{b^{2/3}{a}^{1/3}}{2^{2/3}}+b^{2/3}(2a{)}^{1/3}-c\end{array}$

But, $f\left(x\right)\ge 0$ for all $x>0$

$\begin{array}{l}\therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\frac{b^{2/3}{a}^{1/3}}{2^{2/3}}+b^{2/3}(2a{)}^{1/3}-c\ge 0\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\left(b^{2}a\right)}^{1/3}\left(\frac{1}{2^{2/3}}+2^{1/3}\right)\ge c\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}3{\left(b^{2}a\right)}^{1/3}\ge 2^{2/3}c\Rightarrow 27a{b}^2\ge 4c^3\end{array}$