Problem 7.8.1. Real numbers $a, b, c$ are such that

$$
\left\{\begin{array}{l}
\frac{a b}{a+b}=2 \\
\frac{b c}{b+c}=5 \\
\frac{c a}{c+a}=9
\end{array}\right.
$$

Find the value of the expression $\frac{a b c}{a b+b c+c a}$.
Answer: $\frac{180}{73}$.

Solution. Note that

$$
\begin{gathered}
\frac{1}{a}+\frac{1}{b}=\frac{a+b}{a b}, \quad \frac{1}{b}+\frac{1}{c}=\frac{b+c}{b c}, \quad \frac{1}{c}+\frac{1}{a}=\frac{c+a}{c a} \\
\text { and } \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{a b+b c+c a}{a b c}
\end{gathered}
$$

This means that if we replace the given equalities with equalities of their reciprocals, we get

$$
\left\{\begin{array}{l}
\frac{1}{a}+\frac{1}{b}=\frac{1}{2} \\
\frac{1}{b}+\frac{1}{c}=\frac{1}{5} \\
\frac{1}{c}+\frac{1}{a}=\frac{1}{9}
\end{array}\right.
$$

The value we need to find is the reciprocal of

$$
\begin{aligned}
\frac{1}{a}+\frac{1}{b}+\frac{1}{c} & =\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{b}+\frac{1}{c}+\frac{1}{c}+\frac{1}{a}\right)= \\
& =\frac{1}{2}\left(\frac{1}{2}+\frac{1}{5}+\frac{1}{9}\right)=\frac{45+18+10}{2 \cdot 90}=\frac{73}{180}
\end{aligned}
$$