Baltic Way 1993 · Problem 10

Solution: Let $m$ be such index that $\left| a_{m}^{\prime} - b_{m}^{\prime} \right| = \max_{1 \leq i \leq n} \left| a_{i}^{\prime} - b_{i}^{\prime} \right| = c$. Without loss of generality we may assume $a_{m}^{\prime} > b_{m}^{\prime}$. Consider the numbers $a_{m}^{\prime}, a_{m+1}^{\prime}, \ldots, a_{n}^{\prime}$ and $b_{1}^{\prime}, b_{2}^{\prime}, \ldots, b_{m}^{\prime}$. As there are $n+1$ numbers altogether and only $n$ places in the initial sequence there must exist an index $j$ such that we have $a_{j}$ among $a_{m}^{\prime}, a_{m+1}^{\prime}, \ldots, a_{n}^{\prime}$ and $b_{j}$ among $b_{1}^{\prime}, b_{2}^{\prime}, \ldots, b_{m}^{\prime}$. Now, as $b_{j} \leq b_{m}^{\prime} < a_{m}^{\prime} \leq a_{j}$ we have $\left| a_{j} - b_{j} \right| \geq \left| a_{m}^{\prime} - b_{m}^{\prime} \right| = c$ and $\max_{1 \leq i \leq n} \left| a_{i} - b_{i} \right| \geq c = \max_{1 \leq i \leq n} \left| a_{i}^{\prime} - b_{i}^{\prime} \right|$.