<s>[INST] If $a+b+c=0$ what is the value of $\frac{a^2}{2a^2 +bc }+\frac{b^2}{2b^2 +ca }+\frac{c^2}{2c^2 +ab }$ Let  $s=\frac{a^2}{2a^2 +bc }+\frac{b^2}{2b^2 +ca }+\frac{c^2}{2c^2 +ab }$.
If we use inequality $\frac{x^2}{a}+\frac{y^2}{b} \ge \frac{(x+y)^2}{(a+b)}$ we get $s \ge 0$ as $a+b+c=0$.
Again $s \le \frac{a^2}{bc }+\frac{b^2}{ca }+\frac{c^2}{ab }=\frac{a^3}{abc }+\frac{b^3}{abc }+\frac{c^3}{abc }=3$ as $a+b+c=0$ implies $a^3+b^3+c^3=3abc$.
Thus $0 \le s \le 3$. Putting any one of $a$,$b$,$c$=0 we get $s=1$. Hence my hunch is $s=1$, but i can't prove it.  Can anyone help?
 [/INST] From your observation, we have
$$\sum_{cyc}\frac{a^2}{2a^2+bc}-1=\frac{abc(a+b+c)(ab+bc+ca-a^2-b^2-c^2)}{(2a^2+bc)(2b^2+ac)(2c^2+ab)}$$
Since $a+b+c=0$, we get the value of $S=1$
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