<s>[INST] If $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=1$, what can we say about $\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}$? 
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!
 [/INST] 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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