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.

(In reply to re: Spoiler by Osi)

The proof?  Sure.

ab+bc>cd>ab>bc

Note that all three terms are positive integers.  ab+bc is at least one more than cd, which is at least one more than ab. Therefore, ab+bc-2 is greater than or equal to ab.  Simplifying, bc is greater or equal to 2.  Knowing this, bc=2, ab=3, and cd=4 is the theoretical minimum.  It satisfies the inequalities, so it is the actual minimum.

Posted by Tristan on 2004-11-07 15:19:12