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I'm stuck at this algebra problem, it seems to me that's what's provided doesn't even at all.

Provided: $$a+b+c=0$$

Find the value of: $$\frac{a^2}{2a^2+bc}+\frac{b^2}{2b^2+ac}+\frac{c^2}{2c^2+ab}$$

Like I'm not sure where to start, and the provided clue doesn't even make sense. There's no way I can think of to factor this big polynomial into a form like $a\times p+b\times q+c\times r=s$ where $p,q,r,s\in\mathbb{Z}$.

Thanks!

Spacy
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    do you mean minimum/maximum value? –  May 16 '13 at 18:09
  • What happens when you put $b=c$? What does that tell you about this expression? And if you put $a=0$? – Mark Bennet May 16 '13 at 18:17
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    yes, but I thought we shouldn't just assign variables to a given equation unless you can prove that the polynomial above always give the same value – Spacy May 16 '13 at 18:19
  • But it can give you a good idea what the solution you're looking for might look like. Assigning values or looking at special cases is often a good way to build up some intuition, especially if you're stuck with a problem. – Elmar Zander May 16 '13 at 18:46

2 Answers2

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Note that: $$\frac{a^2}{2a^2+bc}=\frac{a^2}{a^2+a(-b-c)+bc}=\frac{a^2}{(a-b)(a-c)}$$ This applies in the same way for: $$\frac{b^2}{2b^2+ac}=\frac{b^2}{(b-c)(b-a)}\ \text{and}\ \frac{c^2}{2c^2+ab}=\frac{c^2}{(c-a)(c-b)}$$Therefore, the original equation is equal to: \begin{align} &\frac{a^2}{(a-b)(a-c)}+\frac{b^2}{(b-c)(b-a)}+\frac{c^2}{(c-a)(c-b)} \\ \\ &=\frac{-a^2(b-c)-b^2(c-a)-c^2(a-b)}{(a-b)(b-c)(c-a)} \\ \\ &=\frac{-a^2b-b^2c-c^2a+a^2c+b^2a+c^2b}{(a-b)(b-c)(c-a)} \\ \\ &=\frac{(a-b)(b-c)(c-a)}{(a-b)(b-c)(c-a)} \\ \\ &=\boxed{1} \end{align}

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$$ −a^2b−b^2c−c^2a+a^2c+b^2a+c^2b \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \overset{\tiny{+abc-abc}}{=} \Big[ abc+bc^2-b^2c-ac^2 \Big] + \Big[ -a^2b-abc+ab^2+a^2c \Big] \\ =c(ab+bc-b^2-ac)-a(ab+bc-b^2-ac) \\ =(c-a)(ab+bc-b^2-ac) \\ =(c-a)(a-b)(b-c) . $$

Davood
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Ashwin Tomer
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