Ta có: $(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c})^2=\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+\dfrac{2}{ab}+\dfrac{2}{bc}+\dfrac{2}{ca}=\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+\dfrac{2(a+b+c)}{abc}=\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}
\\\Rightarrow \sqrt{\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}}=\sqrt{(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c})^2}=\left | \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c} \right |$