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?
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So the question boils down to showing that A is a perfect square. Am I right? – Qwerty Jun 06 '16 at 20:53
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@AhmedHussein Yes. Sorry I missed the square.. You are right – Qwerty Jun 06 '16 at 21:00
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$$\sum_{cyc}\frac{a^2}{(b-c)^2}-2=\frac{\left(a^3-a^2 b-a b^2+b^3-a^2 c+3 a b c-b^2 c-a c^2-b c^2+c^3\right)^2}{(a-b)^2 (a-c)^2 (b-c)^2}$$ 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.
Jack D'Aurizio
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The cyclic sum is homogeneous of degree zero, so there are either zero solutions or infinitely many. Since there is at least one, there are infinitely many. – Eric Towers Jun 06 '16 at 21:22
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what you wrote on wikipedia seems acceptable, I didn't get there was a two line proof. why don't you write it from the beginning, HERE ? – reuns Jun 06 '16 at 21:53