Prove that $ A=\sqrt{\frac{(b-c)^2}{a^2}+\frac{(c-a)^2}{b^2}+\frac{(a-b)^2}{c^2}} $ is also rational number. Let $a,b,c\in\mathbb{Q}$ distinct and none of them equal to $0$ satisfying $\frac{a^2}{(b-c)^2}+\frac{b^2}{(c-a)^2}+\frac{c^2}{(a-b)^2}\leq 2. $ Prove that $ A=\sqrt{\frac{(b-c)^2}{a^2}+\frac{(c-a)^2}{b^2}+\frac{(a-b)^2}{c^2}} $ is also rational number. Is there a simply way?

$$\sum_{cyc}\frac{a^2}{(b-c)^2}-2=\frac{\left(a^3-a^2 b-a b^2+b^3-a^2 c+3 a b c-b^2 c-a c^2-b c^2+c^3\right)^2}{(a-b)^2 (a-c)^2 (b-c)^2}$$
It follows that the LHS can be $\leq 0$ in very few cases:
$$ abc=(a+b-c)(a-b+c)(-a+b+c).$$
Then it is not difficult to prove that such identity grants $\sum_{cyc}b^2 c^2 (b-c)^2$ to be a rational square.
