## Tuesday, April 10, 2012

### Inequality of the Means

The main problem in this article, the geometric version of the Inequality of the Means, has been chosen for its simplicity and elegance. It is one of those mathematical problems that is easy to state but ridiculously hard to prove. There are two puzzles this month. Please be sure to use the two different Subject lines mentioned in the article to help us distinguish which puzzle it is you are replying to.
INEQUALITY OF THE MEANS
This article’s material came forth from the fertile mind of the extraordinary John Conway. References to the mathematics of the problem are listed at the end. It was a pleasure to discuss this problem with Prof. Dror Bar-Natan. Prof. Bar-Natan’s exposition and Javascript applet make the subject come alive on his website (http://www.math.toronto.edu/~drorbn/), which is certainly worth a visit.
We start with a well-known inequality from high school algebra: let a and b be any two non-negative numbers. Then, their arithmetic mean is at least as large as their geometric mean, i.e.,
square_root(a *b ) <= ((a + b)/2)
Equality occurs precisely when a = b. This is not hard to prove algebraically, but here is a nice geometric proof. Consider the following figure where a square of side length a + b encloses 8 right angled triangles of orthogonal sides a and b each.

The area of the inner square is (a - b)*(a - b), which is the area of the outer square minus the area of the triangles. In other words, we have the following:
(a + b) * (a + b)  - 8. ab. (0.5) = (a - b) * (a - b) >= 0
=> (a + b) * (a + b) >= 4.a.b
=> ((a + b)/2) >= square_root(ab)
The equality discussion is also clear since the inner square has area zero precisely when a = b. The inequality is true for any finite set of non-negative numbers. Given any set of non-negative numbers a1, a2, ..., an, we always have
arithmetic_mean(a1, a2, ..., an) >= geometric_mean(a1, a2, ..., an)
This is not hard to prove algebraically but the fun starts when we try to replicate the geometric argument above. Here is the geometric problem for n = 3: given a cubical box of side length (a + b + c), can you fill it with 27 little cuboids each of length a, breadth b and width c without spilling out of the box? In concrete terms, take a cubical box of size 15 × 15 ×15 (inches for instance) and fill it with 27 little cuboids, each of size 4 × 5 × 6.
The problem is hard, ridiculously hard even. Even if you look at the solution it does not easily lend itself to memory. Here is a link to the solution worked out step by step:
http://www.math.toronto.edu/drorbn/projects/ArithGeom/
Scroll down to the Javascript application or read Professor Bar-Natan’s lucid explanation of the same topic.
For larger n the problem comes in two flavors: solvable or hopeless. That is, if you can solve the problem for m and n, it is possible to bootstrap the solution for nm. Thus, there is a solution for n = 4; 6; 8 etc. But the case n = 5 is wide open. If you have some spare time, then by all means try to fit 3125 five dimensional cuboids of size (a x b x c x d x e) into a five dimensional cube of side length (a + b + c + d + e).
We end this month’s column with a question unrelated to the above inequality. This is the first puzzle for this month.
(New Dice) Take two standard six sided dice numbered 1 to 6 each. Is it possible to number them differently using positive integers so that one gets the same sums (2 through 12) with the same probabilities?