20. Let $a, b$ and $c$ be real numbers such that $\frac{a b}{a+b}=\frac{1}{3}, \frac{b c}{b+c}=\frac{1}{4}$ and $\frac{a a}{c+a}=\frac{1}{5}$. Find the value of $\frac{24 a b c}{a b+b c+c a}$.
20. Answer: 4
Taking reciprocals, we find that $\frac{1}{a}+\frac{1}{b}=3, \frac{1}{b}+\frac{1}{c}=4$ and $\frac{1}{a}+\frac{1}{c}=5$. Summing the three equations, we get
$$
12=2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=2 \times \frac{a b+b c+c a}{a b c} .
$$

Hence $\frac{24 a b c}{a b+b c+c a}=4$.