Example 7 Given that $a, b, c$ are real numbers, and $\frac{ab}{a+b} = \frac{1}{3}, \frac{bc}{b+c} = \frac{1}{4}, \frac{ac}{c+a} = \frac{1}{5}$. Then, the value of $\frac{abc}{ab+bc+ca}$ is $\qquad$
(1997, Hope Cup Mathematics Invitational for Junior High School Second Round)
Solving: Transform the given equations into
$$
\frac{a+b}{a b}=3, \frac{b+c}{b c}=4, \frac{a+c}{a c}=5 \text {. }
$$

Then $\frac{1}{a}+\frac{1}{b}=3, \frac{1}{b}+\frac{1}{c}=4, \frac{1}{a}+\frac{1}{c}=5$.
Adding the three equations yields $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=6$.
Thus, $\frac{a b+b c+a c}{a b c}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=6$,
which means $\frac{a b c}{a b+b c+c a}=\frac{1}{6}$.
In this problem, the numerator and denominator are inverted, making the solution straightforward. The next example involves rearranging the order, using overall transformation to cleverly solve a complex problem.