I found this problem on Cut The Knot, who got it from [1].
You are given seven distinct positive integers that sum to 100. Prove that some three of them must add up to at 50 or more.
I'll put my solution in the comments later this evening.
[1] Andreescu, T. and Răzvan, G. Mathematical Olympiad Challeges(Burkhäuser, 2004, p 60).
Comments
Solution
Suppose we are given seven distinct positive integers that sum to and no three of them sum to or above.
Consider the three largest of the seven integers, since these must sum to more than any other set of three, and suppose they are and in that order. Since they must sum to less than we have .
How large can be? Well it cannot be because the least and could be is and respectively, but which is impossible if .
So is at most and the remaining four numbers of the seven must be less than , so the most they can add up to is . But now the seven numbers cannot add up to . because the original three numbers only added up to at most and which is less than .
So we conclude that for the seven numbers to total there must be some three that add up to or more.