Taking the numerator mod $a,b,c$, we get that $b+c,c+a,a+b$ are multiples of $a,b,c$, respectively. By Taiwan TST 2014, round 3, quiz 3, problem 1, we either have all three variables be equal or the greatest term be the sum of the other two variables. The former case gives $a=b=c=1$, which makes the fraction equal to 8. In the latter case, suppose WLOG that $a+b=c$, then we have that $2b$ is a multiple of $a$ and $2a$ is a multiple of $b$, which forces $b=a$ or $\max(a,b)=2\min(a,b)$. These give $(a,b,c)=(1,1,2),(a,b,c)=(1,2,3)$ and permutations, and $\frac{(a+b)(b+c)(c+a)}{abc}$ is equal to 9 and 10, respectively, in these two cases. Our final answers are $\boxed{8,9,10}$.