If `a x^2+b/xgeqc` for all positive `x` where `a >0` and `b >0,` show that `27 a b^2geq4c^3dot`

Explanation

We have
`ax^(2)+(b)/(x)-cge0,forall,xgt0`
So least value of `f(x)=ax^(2)+(b)/(x)-c` is non negative
Now `f(x)=2ax-(b)/(x^(2))`
`f(x)=0rarrx^(3)=b//2ararrx=(b)/(2a)^(1//3)`
Also `f''(x)=2a+(2b)/(x^(3))`
`f(b)/(2a)^(1//3)=2a+(2b)/(b)xx2a=6agt0`
`therefore f is in at x=((b)/(2a))^(1//3)`
Since (1) is true `forall_(x)` we must have
`f((b)/(2a))^(1//3)ge0`
`(ab)/(2a)+(b)/(b//2a)^(1//3)gec`
`(3b)/(2)(2a)/(b)^(1//3)gec`
`3b^(2//3)(2a)^(1//3)gec`
`(27b^(2))/(8).2agec^(3)`
`27ab^(2)ge4c^(3)`