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Suppose that $a,b,c$ are three real numbers such that $\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}=1$. What are the possible values for $\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}$?

After clearing the denominators, we have $$a(c+a)(a+b)+b(b+c)(a+b)+c(b+c)(c+a)=(a+b)(b+c)(a+c)\,.$$ That is, $$a^3+b^3+c^3+abc=0\,.$$ But then I'm stuck. This question is related, but a bit different.

Thank you for your help!

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    $\begingroup$ What is your source of these problems? Even the last one was brilliant about the triangle. $\endgroup$ – Teresa Lisbon Aug 23 '16 at 13:05
  • $\begingroup$ @астонвіллаолофмэллбэрг: I don't know the exact source of the two problems. These were told to me by a friend of mine. However, one of my previous question was apparently taken from "Aufgabe 11" here. $\endgroup$ – Alphonse Aug 23 '16 at 13:10
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Since $\sum\limits_\text{cyc}\,\frac{a}{b+c}=1$, we have $$a+b+c=(a+b+c)\,\left(\sum_\text{cyc}\,\frac{a}{b+c}\right)=\sum_{\text{cyc}}\left(\frac{a^2}{b+c}+a\right)\,.$$ That is, $$a+b+c=\sum_{\text{cyc}}\left(\frac{a^2}{b+c}\right)+(a+b+c)\,.$$ Hence, $$\sum_{\text{cyc}}\left(\frac{a^2}{b+c}\right)=0\,.$$

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    $\begingroup$ I'm sorry, I don't get it. I get $(a+b+c)\sum_{\text{cyc}}\frac{a}{b+c}=\sum_{\text{cyc}}\left(\frac{a^2}{b+c}+a\left(\frac{b}{c+a}+\frac{c}{a+b} \right) \right)$. How is $\sum_{\text{cyc}}a\left(\frac{b}{c+a}+\frac{c}{a+b} \right) = \sum_{\text{cyc}}a$? Thanks. $\endgroup$ – Bobson Dugnutt Aug 23 '16 at 13:36
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    $\begingroup$ $\frac{a(a+b+c)}{b+c}=\frac{a^2}{b+c}+a$ $\endgroup$ – Batominovski Aug 23 '16 at 13:40
  • $\begingroup$ Thanks! .. it's a neat feature that $(a+b+c)=(b+c+a)=(c+a+b)$, so you can simply include it under the summation. $\endgroup$ – Bobson Dugnutt Aug 23 '16 at 13:45
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Hint. You have

$$ \frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b} - 1 = \frac{a^3+b^3+c^3+abc}{(a+b)(b+c)(c+a)} $$

as well as

$$ \frac{a^2}{b+c} + \frac{b^2}{c+a} + \frac{c^2}{a+b} = \frac{(a^3+b^3+c^3+abc)(a+b+c)}{(a+b)(b+c)(c+a)}. $$

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Assuming

$$\frac a{b+c}+\frac b{c+a}+\frac c{a+b}=1$$

we also have

$$\frac{a^{2}}{b+c}+\frac {ab}{c+a}+\frac {ac}{a+b}= a$$

as well as

$$\frac{ab}{b+c}+\frac {b^2}{c+a}+\frac {bc}{a+b}= b$$

and

$$\frac{ac}{b+c}+\frac {bc}{c+a}+\frac {c^2}{a+b}= c$$

These three sum together as but all terms without $(.)^2$ on the other side on sorting terms with same denominator you get $$\frac{a^{2}}{b+c} + \frac {b^2}{c+a} + \frac {c^2}{a+b} =$$ $$ a+b+c -(\frac {ac}{b+c} + \frac{ab}{b+c}) - (\frac {ab}{c+a} + \frac {bc}{c+a}) - (\frac {ac}{a+b} + \frac {bc}{a+b}) = $$ $$ a + b + c - (a) - (b)- (c) = 0$$

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