P/s: Lâu ngày vào làm bài :v
Ta có: \(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=1\Rightarrow\left(a+b+c\right)\left(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\right)=a+b+c\)
\(\Rightarrow\frac{\left(a+b+c\right)a}{b+c}+\frac{\left(a+b+c\right)b}{a+c}+\frac{\left(a+b+c\right)c}{a+b}=\frac{a^2+a\left(b+c\right)}{b+c}+\frac{b^2+b\left(a+c\right)}{a+c}+\frac{c^2+c\left(a+b\right)}{a+b}\)
\(=\frac{a^2}{b+c}+a+\frac{b^2}{a+c}+b+\frac{c^2}{a+b}=a+b+c\Rightarrow\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b}=0\left(đpcm\right)\)