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.
I've found A = 10, B = 7, C = 5, and D = 1, but I haven't proved yet that this is the lowest.
Clearly A > B > C > D. Therefore, D must equal 1, in the
lowest solution. If it were not, then a lower solution could be
obtained by subtracting (D-1) from A, B, C, D.
I am clear that C cannot equal 2, because C + C > B + D, so C = 2 leaves no acceptable value of B.
Also C cannot equal 3, because C + C > B + D then makes B = 4.
However, A must be less than B + C - 1, and greater than both 2B - C
and 2C - 1, and this doesn't work if C = 3 and B = 4, because A then
must be less than 6 and greater than 5. No room.
If C = 4, then C + C > B + D makes B = 5 or 6. Neither value of B
allows A to be both less than B + C - 1, and greater than both 2B - C
and 2C - 1.
So 5 is the lowest possible value of C.
That's about as far as I've gotten.
A possible solution
