\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\Rightarrow1+a\left(\dfrac{b+c}{bc}\right)=0\Rightarrow a\left(b+c\right)=-bc\left(1\right)\)
\(\left(a-b\right)\left(a-c\right)=a^2-a\left(b+c\right)+bc=a^2+bc+bc=a^2+2bc\left(do.\left(1\right)\right)\)
\(\Rightarrowđpcm\)
Tương tự ta cũng có :
\(\left(b-a\right)\left(b-c\right)=b^2+2ac\)
\(\left(c-b\right)\left(c-a\right)=c^2+2ab\)
\(\Rightarrow\dfrac{bc}{a^2+2bc}+\dfrac{ca}{b^2+2ac}+\dfrac{ab}{c^2+2ab}=\dfrac{bc}{\left(a-b\right)\left(a-c\right)}+\dfrac{ca}{\left(b-c\right)\left(b-a\right)}+\dfrac{ab}{\left(c-a\right)\left(c-b\right)}\)
\(=\dfrac{1}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\left[-bc\left(b-c\right)-ca\left(a-c\right)-ab\left(a-b\right)\right]\) \(\)
\(=\dfrac{1}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\left[a\left(b^2-c^2\right)+b\left(a^2-c^2\right)+c\left(a^2-b^2\right)\right]\) \(\left(do.\left(1\right)\right)\)
\(=\dfrac{1}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}.\left(a-b\right)\left(b-c\right)\left(c-a\right)=1\)
\(\Rightarrowđpcm\)
