$解： ∵\frac {a b}{a+b}=\frac {1}{3}，$
$ ∴a，b均不为 0， 且 \frac {a+b}{a b}=3 . $
$∴\frac {a}{a b}+\frac {b}{a b}= 3，即 \frac {1}{b}+\frac {1}{a}=3①.$
$同理可得 \frac {1}{c}+\frac {1}{b}=4②，\frac {1}{a}+\frac {1}{c}=5③. $
$由①+②+③，得 2(\frac {1}{a}+\frac {1}{b}+\frac {1}{c})=12$
$∴\frac {1}{a}+\frac {1}{b}+\frac {1}{c}=6，即 \frac {a b+b c+a c}{a b c}=6$
$ ∴\frac {a b c}{a b+b c+a c}=\frac {1}{6}$
