$解：∵ \frac{ab}{a+b} = \frac{1}{6}，\frac{bc}{b+c} = \frac{1}{8}，\frac{ac}{a+c} = \frac{1}{10}$
$∴ \frac{a+b}{ab} =6，\frac{b+c}{bc} =8，\frac{a+c}{ac} =10$
$∴ \frac{a+b}{ab} + \frac{b+c}{bc} + \frac{a+c}{ac} =6+8+10$
$∴ \frac{c(a+b)}{abc} + \frac{a(b+c)}{abc} + \frac{b(a+c)}{abc} =24$
$∴ \frac{ac+bc+ab+ac+ba+bc}{abc} =24$
$∴ \frac{2(ab+bc+ac)}{abc} =24$
$\frac{ab+bc+ac}{abc} =12$
$∴ \frac{abc}{ab+bc+ac} =\frac{1}{12}$
