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When a=b=c, the first part of the expression becomes 3 and the second part becomes 9, making 12. Consider the set of numbers where a<b<c, which will be the general case, because the numbers can always be ordered by size. We can write b=a+x and c=a+y, where x and y are not zero or negative, and y>x. The first part of the expression becomes (3a^3+3a^2(x+y)+3a(x^2+y^2)+x^3+y^3)/(a^3+a^2(x+y)+axy).We can divide to get 3 plus a remainder which is non-zero (3a(x^2+y^2)+x^3+y^3-3axy), because (3a(x^2+y^2)+x^3+y^3)>3axy or (x^2+y^2)+(x^3+y^3)/3a>xy. We can rewrite this expression as x^3/3a+x^2-xy+y^3/3a+y^2>0, or x^2(x/3a+1)>-y(y^2/3a+y-x). Clearly the right side of this inequality is always negative, because x and y are both positive and y>x by definition. The inequality is always true. This proves that the remainder of the earlier division is positive so the first term is greater than or equal to 3.

We can approach the second term similarly.

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