Problem 6. We will "flip" each of the equations in the system:

$$
\left\{\begin{array}{l}
\frac{a+b}{a b}=1 \\
\frac{b+c}{b c}=\frac{1}{2} \\
\frac{c+a}{c a}=\frac{1}{4}
\end{array}\right.
$$

(The transformation is equivalent since none of the right-hand sides can become zero).

Notice that $\frac{a+b}{a b}=\frac{1}{a}+\frac{1}{b}$, and so on. Therefore, we have obtained a system of linear equations in $\frac{1}{a}, \frac{1}{b}$, and $\frac{1}{c}$.

Solving it, we get $\frac{1}{a}=\frac{3}{8}, \frac{1}{b}=\frac{5}{8}, \frac{1}{c}=-\frac{1}{8}$.
Answer. $a=\frac{8}{3}, b=\frac{8}{5}, c \stackrel{=}{=}-8$.