Problem 1
Let be distinct elements of such that whenever (with ), the sum is also one of the . Prove .
Step 2 of 3: Prove the paired-sum inequality
In plain words
Prove the paired-sum inequality
Detailed analysis
Suppose instead that . Then the distinct sums for are all at most , so each belongs to the set. Every one is strictly larger than , but there are only elements with indices , a contradiction.