2026/08/30

$abc \ne 0, \dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}=1$, find $\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}$

Solve: \(\begin{multline} \shoveleft \text{Let }A=\dfrac{a}{b+c}, B=\dfrac{b}{c+a}, C=\dfrac{c}{a+b}, D=Aa+Bb+Cc\\ \shoveleft \implies A+B+C=1, a=Ab+Ac, b=Bc+Ba, c=Ca+Cb\\ \shoveleft \implies a+b+c=(1-A)a+(1-B)b+(1-C)c=a+b+c-D\\ \shoveleft \implies D=\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}=\bbox[5px, border: 1px solid black]{0} \end{multline}\)