Ta có: \(\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}=1\)
=> \(\left(a+b+c\right)\left(\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}\right)=a+b+c\)
<=> \(\dfrac{a^2}{b+c}+\dfrac{ab}{b+c}+\dfrac{ac}{b+c}+\dfrac{b^2}{a+c}+\dfrac{ab}{a+c}+\dfrac{bc}{a+c}+\dfrac{c^2}{a+b}+\dfrac{ac}{a+b}+\dfrac{bc}{a+b}=a+b+c\)
<=> \(\dfrac{a^2}{b+c}+\dfrac{b^2}{a+c}+\dfrac{c^2}{a+b}+a\left(\dfrac{b}{b+c}+\dfrac{c}{b+c}\right)+b\left(\dfrac{a}{a+c}+\dfrac{c}{a+c}\right)+c\left(\dfrac{a}{a+b}+\dfrac{b}{a+b}\right)=a+b+c\)
<=> \(\dfrac{a^2}{b+c}+\dfrac{b^2}{a+c}+\dfrac{c^2}{a+b}+a+b+c=a+b+c\)
<=> \(\dfrac{a^2}{b+c}+\dfrac{b^2}{a+c}+\dfrac{c^2}{a+b}=0\)
Vậy \(\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}=1\) thì \(\dfrac{a^2}{b+c}+\dfrac{b^2}{a+c}+\dfrac{c^2}{a+b}=0\)
