1. Given $\frac{a b}{a+b}=\frac{1}{15}, \frac{b c}{b+c}=\frac{1}{17}, \frac{c a}{c+a}=\frac{1}{16}$. Then the value of $\frac{a b c}{a b+b c+c a}$ is ( ).
(A) $\frac{1}{21}$
(B) $\frac{1}{22}$
(C) $\frac{1}{23}$
(D) $\frac{1}{24}$
1.D.

From the given, we have
$$
\frac{1}{a}+\frac{1}{b}=15, \frac{1}{b}+\frac{1}{c}=17, \frac{1}{c}+\frac{1}{a}=16 \text {. }
$$

Adding the three equations yields $2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=48$.
Therefore, the original expression $=\frac{1}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}=\frac{1}{24}$.