I'm thinking of four positive integers A, B, C, and D. A>B>C>D is true.

I have written down another inequality that is also true, but I'm not showing it to you. This inequality puts the following values in order from greatest to smallest:
A, A+C, B, A+D, C, B+C, D, B+D, A+B, C+D
(this is obviously not the order, as A+C can't be less than A or less than C)

I showed the two inequalities to my friend, and he was able to minimize A, B, C, and D all at once. I told him that he had just guessed correctly which numbers were on my mind.

Based only on this information, what is the highest possible sum of the four numbers on my mind?

(In reply to re: a start + Possible solution by Hugo)

"a+b > a+c > a+d > a > b+c > b+d > b > c+d > c > d
Where a=10, b = 6, c = 3, d = 1"

Mmm, I disagree Hugo.  If I had a=500, b=6, c=3, d=1 those inequalities would still be true.  So a hasn't been minimized yet.

I could be misunderstanding that part of the puzzle, though.  I understood it to mean that "based on the inequality, my friend was able to say A£w, B£x, C£y, and D£z.  As it turns out, w,x,y,z (the maximum possible values) were in fact the numbers I was thinking of."  And then the highest possible sum of the 4 numbers that satisfy the inequality is w+x+y+z.

If my understanding is correct, then the only inequality could be
A+B > A+C > B+C > A+D > A > B+D > C+D > B > C > D

Here's how I got there:

Mini inequalities we know are true are:
A+B>A+C>A+D>A>B>C>D
B+C>B+D>B>C>D
C+D>C>D
A+C>B+C>B+D>C+D

This means that B+C could greater than A+D, or less than A, or between A+D and A.

Now B+D could not be greater than A+D, but it could be greater than A or less than A.

C+D can¡¯t be greater than B+D but it could be greater than A or less than B or anywhere in-between (sort of).

If we had A+B>A+C>A+D>"everything else", then we wouldn¡¯t have any "cap" on what A could be. So I think the inequality must be A+B>A+C>B+C>A+D>"everything else". I know that if we had A+B > A+C > B+C > A+D > B+D > C+D > C+D > A > B > C > D then there would be no limit on any of the numbers because A,B,C,D = (100,99,98,96 or 200,199,198,196) both would fit into these inequalities, and you can see I could just keep making them bigger.

So I think the only 4 inequalities left to consider are:

A+B > A+C > B+C > A+D > B+D > A > C+D > B > C > D
A+B > A+C > B+C > A+D > B+D > A > B > C+D > C > D
A+B > A+C > B+C > A+D > A > B+D > C+D > B > C > D
A+B > A+C > B+C > A+D > A > B+D > B > C+D > C > D

I think the first step is to see if B+D is greater than or less than A. This should be done in a way to restrict B. Having A+B > A+C > B+C > A+D > B+D > everything else does not restrict B. As long as A and B grow together, they could become as great as they want. So we must have A+B > A+C > B+C > A+D > A > B+D > everything else.

This in itself does not restrict A and B. Technically in this arrangement they could still grow together. However, B does not have as much freedom to get as close to A, so I believe this is a more restricting scenario.

I know that if we have everything else > B > C+D > C > D that A and B can grow under my A+B > A+C > B+C > A+D > A > B+D > everything else rule. I found this out by just playing around with numbers. So it must be that:

A+B > A+C > B+C > A+D > A > B+D > C+D > B > C > D

But I don't know what to do with that yet to find w,x,y,z.  Hrmm

Edited on January 13, 2005, 9:02 pm

Posted by nikki on 2005-01-13 20:40:41