When is this an Integer?

This problem has a very simple application of the famous AM-GM inequality. Given expression, ((ax^2+bx)(bx^2+cx)(cx^2+ax)/abc)^(1/3), can be simplified to, x(x^3+1+x^2(a/b+b/c+c/a)+ x(a/c+c/b+b/a))^(1/3) Now since, a,b,c are positive, coefficients of x and x^2 are >=3, if the vary over positive reals. Thus minimum value is x(x+1). x can vary from 0 to 1000. Thus maximum possible value for minimum expression is 1000 \times 1001 = \boxed{1001000}.

Let's first break this problem into parts. First, we'll find the value of $\text{min}\left[\prod_{cyc}\left(\dfrac{ax^2+bx}{c}\right)\right]$

This is equivalent to finding the largest value of $k$ such that $\prod_{cyc}(ax^2+bx)\ge kabc$ is true.

By Holder's Inequality, $\prod_{cyc}(ax^2+bx)\ge (\sqrt[3]{abc}x^2+\sqrt[3]{abc}x)^3=abc(x^2+x)^3$

So $k=(x^2+x)^3$ and $\text{min}\left[\prod_{cyc}\left(\dfrac{ax^2+bx}{c}\right)\right]=(x^2+x)^3$

But then $\sqrt[3]{\text{min}\left[\prod_{cyc}\left(\dfrac{ax^2+bx}{c}\right)\right]}=x^2+x$

so we just need to know the number of times $x^2+x$ is an integer when $0 < x\le 1000$

Clearly $x^2+x=1$ is the smallest value with $x$ in the range that gives a solution, and $x=1000$ gives $x^2+x=1001000$ which is the largest value with $x$ in the range.

Since $f(x)=x^2+x$ is a continuous strictly increasing function for $x>0$ , then every single value of $x^2+x$ between $1$ and $1001000$ is represented by a unique $x$ between $0\to 1000$ , so the answer is $\boxed{1001000}$ .