Hiển nhiên thấy $a+b+c\ne0$ vì nếu ngược lại $a+b+c=0$ thì
$\frac{a}{b+c} + \frac{b}{a+c} + \frac{c}{a+b} = \frac{a}{-a} + \frac{b}{-b} + \frac{c}{-c} = -3 \ne 1$.
Ta có
$\left ( a+b+c \right )\left ( \frac{a}{b+c} + \frac{b}{a+c} + \frac{c}{a+b}\right )=a+b+c$
$\Rightarrow \left ( a+b+c \right ) \frac{a}{b+c} +\left ( a+b+c \right ) \frac{b}{a+c} +\left ( a+b+c \right ) \frac{c}{a+b} =a+b+c$
$\Rightarrow \frac{a^2}{b+c} + \frac{b^2}{a+c}+ \frac{c^2}{a+b} +a+b+c=a+b+c$
$\Rightarrow \frac{a^2}{b+c} + \frac{b^2}{a+c} +\frac{c^2}{a+b}=0.$