Find the value of $\frac{a^2}{2a^2+bc}+\frac{b^2}{2b^2+ca}+\frac{c^2}{2c^2+ab}$ 
Let $a$, $b$ and $c$ be real numbers such that $a + b + c = 0$ and define:
  $$P=\frac{a^2}{2a^2+bc}+\frac{b^2}{2b^2+ca}+\frac{c^2}{2c^2+ab}.$$
  What is the value of $P$?

This question came in the regional maths olympiad. I tried AM-GM and CS inequality but failed to get a result. Please give me some hint in how to solve this question.  
 A: For $\prod\limits_{cyc}(a-b)\neq0$ we obtain:
$$\sum_{cyc}\frac{a^2}{2a^2+bc}=\sum_{cyc}\frac{a^2}{a(a+b+c)+a^2-ab-ac+bc}=$$
$$=\sum_{cyc}\frac{a^2}{(a-b)(a-c)}=\frac{\sum\limits_{cyc}a^2(c-b)}{\prod\limits_{cyc}(a-b)}=1.$$
A: If $a=3$ and $b=-1$ and $c=-2$ we get a value $1$.
But
$$\sum_{cyc}\frac{a^2}{2a^2+bc}-1=-\frac{abc(a+b+c)\sum\limits_{cyc}(a^2-ab)}{\prod\limits_{cyc}(2a^2+bc)}=0.$$
Thus, $P=1$ for all $a+b+c=0$ and $\prod\limits_{cyc}(2a^2+bc)\neq0$
and we are done!
A: Write $c=-a-b$.
Then
$$\frac{a^2}{2a^2+bc}=\frac{a^2}{2a^2-ab-b^2}=\frac{a^2}{(2a+b)(a-b)}$$
and
$$\frac{c^2}{2c^2+ab}=\frac{(a+b)^2}{2a^2+5ab+2b^2}=\frac{(a+b)^2}{(2a+b)(a+2b)}.$$
The sum therefore equals
$$\frac{a^2(a+2b)-b^2(2a+b)+(a+b)^2(a-b)}{(2a+b)(a+2b)(a-b)}=\cdots$$
etc. (as long as the denominator is nonzero).
A: plugging $c=-a-b$ in your term, we get $${\frac {{a}^{2}}{2\,{a}^{2}+b \left( -a-b \right) }}+{\frac {{b}^{2}}{
 \left( -a-b \right) a+2\,{b}^{2}}}+{\frac { \left( -a-b \right) ^{2}
}{ab+2\, \left( -a-b \right) ^{2}}}
$$
simplifying this we get $1$
