Evaluating $P= \dfrac{a^2}{2a^2+bc}+\dfrac{b^2}{2b^2+ca}+\dfrac{c^2}{2c^2+ab}$

Question

We are given that, $$a+b+c=0$$ Then, what is the value of: $$P= \dfrac{a^2}{2a^2+bc}+\dfrac{b^2}{2b^2+ca}+\dfrac{c^2}{2c^2+ab}$$

Answer 1

Use:$$\sum_{cyc}\frac{a^2}{2a^2+bc}=\sum_{cyc}\frac{a^2}{2a^2+bc-a(a+b+c)}=\sum_{cyc}\frac{a^2}{(a-b)(a-c)}=$$ $$=\sum_{cyc}\frac{a^2(b-c)}{(a-b)(a-c)(b-c)}=\frac{(a-b)(a-c)(b-c)}{(a-b)(a-c)(b-c)}=1.$$