A interesting number theory problem

Without loss of generality, a≥b≥c. Consider that when(a,b,c)=(2,2,2), (3,2,2), (3,3,2), (4,2,2) ,the value of this algebraic expression A are equal to 2, $\frac{3}{2}$ , $\frac{17}{8}$ , $\frac{11}{4}$ . It shows that if a+b+c ≤ 8, A≥ $\frac{3}{2}$ .

Then prove: if a+b+c≥9, A≥ $\frac{3}{2}$ .

In fact, if A≥ $\frac{3}{2}$ , that means a^2+b^2+c^2+2∑（ab-[a,b])≥3(a+b+c)

For any positive integer x and y, we get: xy≥[x,y]

That means we only need to prove: a^2+b^2+c^2≥3(a+b+c)

Because a+b+c≥9, (a-b)^2+(b-bc)^2+(c-a)^2≥0

So 2(a^2+b^2+c^2)-2(ab-bc-ca)≥0

3(a^2+b^2+c^2)≥(a+b+c)^2

a^2+b^2+c^2≥ $\frac{1}{3}$ (a+b+c)^2

It is obviously correct.

So the answer is $\frac{3}{2}$ .