What are the smallest positive integers A, B, C, and D such that A+A > A+B > A+C > B+B > B+C > A+D > C+C > B+D > C+D > D+D ?
Note: Of all solutions, choose the one with the smallest A, then smallest B if there are more than one with the smallest A, etc.
We have
A+A > A+B > A+C > B+B > B+C > A+D > C+C > B+D > C+D > D+D. (0)
The inequalities can be summarized as: A > B > C > D, and a few others. We use some of those others to derive inequalities connecting the differences between successive numbers: A-B, B-C, and C-D.
A+C > B+B implies A-B > B-C (subtract B+C)
B+C > A+D implies C-D > A-B (subtract B+D)
From (0) we also have B+B-(C+C) > 2, so B-C > 1.
Putting these together, we have C-D > A-B > B-C > 1. (1)
Note that (0) is satisfied if we add a constant to each variable. So, without loss of generality, we may set D = 1.
The smallest possible choices from (1) are B-C = 2, A-B = 3, C-D = 4.
This yields (D, C, B, A) = (1, 5, 7, 10), which satisfies all of the original inequalities, and is minimal for D, C, B, and A.
Solution
