Suppose that the integers 9, 10, $\dots$ , 20 are listed in some arbitrary order $(a_1, a_2, \dots)$. Prove that $\exists i, 0 \leq i \leq 10 $ and $a_i + a_{i+1} + a_{i+2} \geq 44$.
This is saying that no matter how you arrange the numbers, in the first 10, you can get three in a row that sum to 44 or more. I assume I would use a prove by contradiction to solve this, but I'm not sure where to start.
Let $s_1$ be the sum of the first three numbers in the list, $s_2$ the sum of the next three, $s_3$ the sum of the next three, and $s_4$ the sum of the last three. Clearly
$$s_1+s_2+s_3+s_4=\sum_{k=9}^{20}k=174\,.$$
The average of these four numbers is $43.5$, so ... ?