Let a,b,c be positive numbers. If $a{x}^2+\frac{b}{x}\ge cforallx>0$ , then the minimum value of $\frac{a{b}^2}{c^3}$  is

$f\left(x\right)=a{x}^2+\frac{b}{x}\Rightarrow f^1\left(x\right)=2ax-\frac{b}{x^2}\Rightarrow f^{11}\left(x\right)=2a+\frac{2b}{x^3}>0$

f(x) is minimum for $f^1\left(x\right)=0,x={\left(\frac{b}{2a}\right)}^{1/3}$

min. $f=a{\left(\frac{b}{2a}\right)}^{2/3}+b{\left(\frac{2a}{b}\right)}^{1/3}={\left(\frac{27a{b}^2}{4}\right)}^{1/3}\ge c\Rightarrow \frac{a{b}^2}{c^3}\ge \frac{4}{27}$