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.
For $\prod\limits_{cyc}(a-b)\neq0$ we obtain: $$\sum_{cyc}\frac{a^2}{2a^2+bc}=\sum_{cyc}\frac{a^2}{a(a+b+c)+a^2-ab-ac+bc}=$$ $$=\sum_{cyc}\frac{a^2}{(a-b)(a-c)}=\frac{\sum\limits_{cyc}a^2(c-b)}{\prod\limits_{cyc}(a-b)}=1.$$
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!
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).
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 $1$
