The 2020 numbers

$$
1 \cdot 2, 2 \cdot 3, 3 \cdot 4, \ldots, 2020 \cdot 2021
$$

are written on the board. A move consists of choosing three numbers \(a, b, c\) written on the board, erasing them, and writing on the board the number

$$
\frac{a b c}{a b + b c + a c}
$$

a) Verify that

$$
\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \left( \frac{a b c}{a b + b c + a c} \right)^{-1}
$$

b) Using

$$
\frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1}
$$

find the value of the sum

$$
\frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 4} + \ldots + \frac{1}{2020 \cdot 2021}
$$

c) After 1009 moves, two numbers remain on the board. Prove that if one of them is \( \frac{4}{3} \), the other is greater than 4.
Solution

a) Note that

$$
\begin{aligned}
\frac{1}{a}+\frac{1}{b}+\frac{1}{c} & =\frac{b c+a c+a b}{a b c} \\
& =\left(\frac{a b c}{a b+b c+a c}\right)^{-1}
\end{aligned}
$$

b) By virtue of the suggested identity, we have

$$
\begin{aligned}
\frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\frac{1}{3 \cdot 4}+\ldots+\frac{1}{2020 \cdot 2021} & = \\
\left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\ldots+\left(\frac{1}{2020}-\frac{1}{2021}\right) & = \\
\frac{1}{1}-\frac{1}{2021} & =\frac{2020}{2021}
\end{aligned}
$$

c) From the previous item, at any moment, the sum of the inverses of the numbers written on the board is always the same. At the end, if only the numbers $x$ and $4 / 3$ remain, we have

$$
\frac{1}{x}+\frac{3}{4}=\frac{2020}{2021}
$$