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let $a,b,c$ such $$\dfrac{a^2}{b+c-a}+\dfrac{b^2}{a+c-b}+\dfrac{c^2}{a+b-c}=0$$

Prove that: $$F(a,b,c)=\dfrac{a}{b+c-a}+\dfrac{b}{a+c-b}+\dfrac{c}{a+b-c}=1 \text{or}-\dfrac{3}{2}$$

maybe can use $$\dfrac{a^2}{b+c-a}+a+\dfrac{b^2}{a+c-b}+b+\dfrac{c^2}{a+b-c}+c=a+b+c$$ $$\Longrightarrow \dfrac{ab+ac}{b+c-a}+\dfrac{ba+bc}{a+c-b}+\dfrac{ac+bc}{a+b-c}=0$$ maybe we can have $a+b+c=0?$

math110
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    How is this a calculus problem? Looks like an algebra problem to me. – Mathemagician1234 Jan 08 '15 at 16:14
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    Do you have a reason to think there is a solution? You have one equation in three unknowns, so one would expect a 2D space of solutions. You are asking that $F(a,b,c)$ be constant over that space. If you trust the problem setter to make sure there is an answer, you can find any solution to the equation and evaluate $F$ – Ross Millikan Jan 08 '15 at 16:18
  • such this problem: http://math.stackexchange.com/questions/581884/how-prove-this-fracabc-a22-fracbca-b22-fraccab-c22-0?rq=1 – math110 Jan 08 '15 at 16:40
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    Problems like this have to be carefully prepared to have an answer. The one you link to was prepared this way. Otherwise the solution space argument applies. – Ross Millikan Jan 08 '15 at 16:50
  • No,This problem have result: it is $1$ or $-\dfrac{3}{2}$ – math110 Jan 09 '15 at 05:24

2 Answers2

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There is no answer. I arbitrarily chose $b=1,c=2$ and solved for $a$. Alpha returned four solutions, among them $-3$ and $1-2^{(2/3)}$. Evaluating your expression gives $1$ for $1-2^{(2/3)}$ and $\frac 32$ for $-3$

Ross Millikan
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hint: the given condition is equivalent to $$- \left( a+b+c \right) \left( {a}^{3}-b{a}^{2}-c{a}^{2}-{b}^{2}a+4\,c ba-{c}^{2}a+{b}^{3}-{b}^{2}c-b{c}^{2}+{c}^{3} \right) =0$$

Dr. Sonnhard Graubner
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