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!
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, $$\sum_{\text{cyc}}\left(\frac{a^2}{b+c}\right)=0\,.$$
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 $$\frac{a^{2}}{b+c} + \frac {b^2}{c+a} + \frac {c^2}{a+b} =$$ $$ a+b+c -(\frac {ac}{b+c} + \frac{ab}{b+c}) - (\frac {ab}{c+a} + \frac {bc}{c+a}) - (\frac {ac}{a+b} + \frac {bc}{a+b}) = $$ $$ a + b + c - (a) - (b)- (c) = 0$$
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
$$ \frac{a^2}{b+c} + \frac{b^2}{c+a} + \frac{c^2}{a+b} = \frac{(a^3+b^3+c^3+abc)(a+b+c)}{(a+b)(b+c)(c+a)}. $$
