Ta có:
\(\left(a+b+c\right)\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)=a+b+c\) (Do \(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=1\) )
\(\Leftrightarrow\frac{a^2}{b+c}+a+\frac{b^2}{c+a}+b+\frac{c^2}{a+b}+c=a+b+c\)
\(\Leftrightarrow\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}=a+b+c-\left(a+b+c\right)=0\left(đpcm\right)\)
*Lưu ý: Có rút gọn một số bước.