Equality of square roots and absolute value

$\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}} = |\frac{1}{a}+\frac{1}{b}+\frac{1}{c} |$

Since both sides of the equation are always positive for every non-zero real $a,b,c$ , we have:

$\Rightarrow (\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}})^2 = (|\frac{1}{a}+\frac{1}{b}+\frac{1}{c} |)^2$

$\Leftrightarrow \frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2} = (\frac{1}{a}+\frac{1}{b}+\frac{1}{c})^2 = \frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2} + 2( \frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca})$

$\Leftrightarrow 2( \frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}) = 0 \Leftrightarrow \frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca} = 0$

$\Leftrightarrow \frac{a+b+c}{abc} = 0$

$\Rightarrow a+b+c=\boxed{0}$

For all nonzero $a,b,c$ , we have:

$|\frac{1}{a} + \frac{1}{b} + \frac{1}{c}| = |\frac{bc+ac+ab}{abc}| = \sqrt{\frac{(bc+ac+ab)^2}{a^2b^2c^2}} = \sqrt{\frac{a^2b^2 + a^2c^2 +b^2c^2 + 2abc(a+b+c)}{a^2b^2c^2}} = \sqrt{\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} + 2(\frac{a+b+c}{abc})} \Rightarrow \boxed{a+b+c=0}.$