Solving $L= \frac{a^2}{2a^2+bc}+\frac{b^2}{2b^2+ac}+\frac{c^2}{2c^2+ab}$ priveded $a+b+c=0$

Question

Let $a,b,c$ be such that $a+b+c=0$ and suppose that $$L= \frac{a^2}{2a^2+bc}+\frac{b^2}{2b^2+ac}+\frac{c^2}{2c^2+ab}.$$ Find the value of $L$.

I can only see the symmetry of these function but cannot solve it.

Answer 1

Since $a+b+c|(a^3+b^3+c^3-3abc)$, $a+b+c = 0 \implies a^3+b^3+c^3 = 3abc$.

Then Wolfram Alpha says that this gives the sum as $1$.

Answer 2

Since $a+b+c=0$ it follows $c=-a-b$, then \begin{align} 2a^2+bc&=2a^2+b(-a-b)\\ &=2a^2-ab-b^2\\ &=(a-b)(2a+b)\\ &=(a-b)(a+a+b)\\ &=(a-b)(a-c)\quad\text{because $a+b=-c$} \end{align} In a similar way we can obtain $2b^2+ac=(b-a)(b-c)$ and $2c^2+ab=(c-a)(c-b)$. Then \begin{align} L&=\frac{a^2}{(a-b)(a-c)}+\frac{b^2}{(b-a)(b-c)}+\frac{c^2}{(c-a)(c-b)} \\ &=\frac{a^2}{(a-b)(a-c)}-\frac{b^2}{(a-b)(b-c)}+\frac{c^2}{(a-c)(b-c)} \\ &=\frac{a^2(b-c)-b^2(a-c)+c^2(a-b)}{(a-b)(a-c)(b-c)} \\ &=\frac{-a(b^2-c^2)+b(a^2-c^2)-c(a^2-b^2)}{(a-b)(a-c)(b-c)} \\ &=\frac{-a(b+c)(b-c)+b(a+c)(a-c)-c(a+b)(a-b)}{(a-b)(a-c)(b-c)} \\ &=\frac{-a(b+c)(b-c)+b(a+c)(a-c)-c(a+b)[(a-c)-(b-c)]}{(a-b)(a-c)(b-c)} \\ &=\frac{[-a(b+c)+c(a+b)](b-c)+[b(a+c)-c(a+b)](a-c)}{(a-b)(a-c)(b-c)} \\ &=\frac{-b(a-c)(b-c)+a(b-c)(a-c)}{(a-b)(a-c)(b-c)} \\ &=\frac{(a-b)(a-c)(b-c)}{(a-b)(a-c)(b-c)} \\ &=1 \end{align}