Xét hiệu : \(\left(\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+a}\right)-\left(\frac{a^2}{a+c}+\frac{b^2}{b+a}+\frac{c^2}{c+b}\right)\)
\(=\frac{a^2}{a+b}-\frac{b^2}{b+a}+\frac{b^2}{b+c}-\frac{c^2}{c+b}+\frac{c^2}{c+a}+\frac{a^2}{a+c}\)
\(=\frac{a^2-b^2}{a+b}+\frac{b^2-c^2}{b+c}+\frac{c^2-a^2}{c+a}\)
\(=\frac{\left(a+b\right)\left(a-b\right)}{a+b}+\frac{\left(b-c\right)\left(b+c\right)}{b+c}+\frac{\left(c+a\right)\left(c-a\right)}{c+a}\)
\(=a-b+b-c+c-a=0\)
Nên: \(\left(\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+a}\right)=\left(\frac{a^2}{a+c}+\frac{b^2}{b+a}+\frac{c^2}{c+b}\right)=\frac{16}{17}\)
Vậy : \(\left(\frac{a^2}{a+c}+\frac{b^2}{b+a}+\frac{c^2}{c+b}\right)=\frac{16}{17}\)
