Ta có :\(\left(\dfrac{1}{a}-\dfrac{1}{b}-\dfrac{1}{c}\right)^2=\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}-\dfrac{2}{ab}+\dfrac{2}{bc}-\dfrac{2}{ac}\)
\(\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}=\left(\dfrac{1}{a}-\dfrac{1}{b}-\dfrac{1}{c}\right)^2+\dfrac{2}{ab}-\dfrac{2}{bc}+\dfrac{2}{ac}\)
\(\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}=1+2\left(\dfrac{c-a+b}{abc}\right)\)
\(\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}=1+2\left(\dfrac{c-\left(a-b\right)}{abc}\right)\left(1\right)\)
Theo đề ra : a=b+c
\(\Leftrightarrow c=a-b\)
\(\Leftrightarrow c-\left(a-b\right)=0\)
\(\left(1\right)\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}=1+2\left(\dfrac{0}{abc}\right)=1\)
\(Hay\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}=1\left(đpcm\right)\)
