Ta có $a+b+c=\frac{1}{abc} \Rightarrow abc(a+b+c)=1$.
Do đó $1+b^2c^2 = abc(a+b+c) + b^2c^2 = bc(a+b)(a+c)$.
Tương tự, ta có $1+a^2c^2 = ac(a+b)(b+c), 1+a^2b^2 = ab(b+c)(a+c)$.
Suy ra
$\sqrt{\frac{(1+b^2c^2)(1+a^2c^2)}{c^2+a^2b^2c^2}} = \sqrt{\frac{(1+b^2c^2)(1+a^2c^2)}{c^2(1+a^2b^2)}} = \sqrt{\frac{bc(a+b)(a+c)ac(a+b)(b+c)}{c^2ab(a+c)(b+c)}} = \sqrt{(a+b)^2} = a+b$ (đpcm).