Problem 1. Let $a, b, c$ be numbers different from zero, such that $b(c+a)$ is the arithmetic mean of the numbers $a(b+c)$ and $c(a+b)$. If $b=\frac{2019}{2020}$, calculate the arithmetic mean of the numbers $\frac{1}{a}, \frac{1}{b}$ and $\frac{1}{c}$.
Solution. From the condition of the problem, we have $2 b(c+a)=a(b+c)+c(a+b)$, which is equivalent to $2 a c=a b+b c$. However, $a b c \neq 0$, so from the last equality, we get $\frac{2}{b}=\frac{1}{a}+\frac{1}{c}$. Using the last equality, the desired arithmetic mean is obtained as

$$
\frac{1}{3}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=\frac{1}{3}\left(\left(\frac{1}{a}+\frac{1}{c}\right)+\frac{1}{b}\right)=\frac{1}{3}\left(\frac{2}{b}+\frac{1}{b}\right)=\frac{1}{b}=\frac{2020}{2019}
$$