Tuesday, April 10, 2012

Inequality of the Means

Indiatimes is running an edited version of this article, which is the next installment of the puzzle column. This article is in collaboration with Prof. Krishnan Shankar of the University of Oklahoma.

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.
This article is in collaboration with Prof. Krishnan Shankar, Professor of Mathematics at the University of Oklahoma.
The Oracle Asks
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.