\(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac})=4 \\<=>\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2.\dfrac{a+b+c}{abc}=4 \\<=>\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2=4(do\ a+b+c=abc) \\<=>\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}=2 (đpcm)\)