Question

Cho \(\frac{a}{b+c}\)+\(\frac{b}{c+a}\)+\(\frac{c}{a+b}\)=1. Chứng minh rằng \(\frac{a^2}{b+c}\)+\(\frac{b^2}{c+a}\)+\(\frac{c^2}{a+b}\)= 0

Answer 1

\(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=1\)

\(\Rightarrow\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)\left(a+b+c\right)=a+b+c\)

\(\Rightarrow\frac{a^2}{b+c}+\frac{a\left(b+c\right)}{b+c}+\frac{b^2}{a+c}+\frac{b\left(a+c\right)}{a+c}+\frac{c^2}{a+b}+\frac{c\left(a+b\right)}{a+b}=a+b+c\)

\(\Rightarrow\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}+a+b+c=a+b+c\)

\(\Rightarrow\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}=0\)

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