AUT_ABooklet_2023

The expression $\frac{x+y}{4}$ reminds us of the arithmetic mean. By the AM-HM inequality, we have $$\frac{x+y}{2}\ge \frac{2}{\frac{1}{x}+\frac{1}{y}}$$ or $$\frac{1}{x}+\frac{1}{y}\ge \frac{1}{(x+y)/4}$$ This inequality leads us to consider an argument concerning the reciprocals of the numbers on the board, as the sum of the reciprocals of two of these numbers is at least as large as the reciprocal of the number replacing them. This value remains the same if and only if the two chosen numbers are equal, and is otherwise larger. At the beginning, the sum of all reciprocals is $$\frac{1}{2023}+\frac{1}{2023}+\cdots +\frac{1}{2023}=\frac{2023}{2023}=1.$$ This implies the claim, since there is an odd number of $2023$s in the beginning that cannot be divided into pairs, so one of them has to be part of a pair with different numbers.