Giải:
Ta có:
\(\dfrac{1}{a^2}+\dfrac{1}{ab}+\dfrac{1}{ac}=\dfrac{2}{a}\)
\(\dfrac{1}{b^2}+\dfrac{1}{ab}+\dfrac{1}{bc}=\dfrac{2}{b}\)
\(\dfrac{1}{c^2}+\dfrac{1}{ac}+\dfrac{1}{bc}=\dfrac{2}{c}\)
Cộng theo vế, ta được:
\(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac}\right)=2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
\(\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac}\right)=2.2\)
\(\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac}\right)=4\)
\(\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2\left(\dfrac{a+b+c}{abc}\right)=4\)
\(\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2=4\)
\(\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}=2\)
Vậy ...
Khai thác giả thiết: \(a+b+c=abc\Leftrightarrow\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac}=1\)(1)
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=2\)
\(\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+\dfrac{2}{ab}+\dfrac{2}{bc}+\dfrac{2}{ac}=4\)
\(\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2=4\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}=2\left(đpcm\right)\)