The harmonic mean of a set of positive numbers is equal to the inverse of the average of the inverses of the numbers. The geometric mean of a set of positive numbers is equal to the nth root of the product of the numbers, where n is the size of the set. The arithmetic mean is equal to the average of the numbers.

It is known that the arithmetic mean is always greater than or equal to the geometric mean, given a set of positive numbers. But where does the harmonic mean fit in with these two other means? Is it greater, lesser, or inbetween? Prove it. Note that in this problem, a "set" allows repeated numbers.

Let a,b,...,z be positive real numbers.  Then (ab...z)^(1/n)<=(a+b+...+z)/n is equivalent to (log(a)+log(b)+...+log(z))/n<=log((a+b+...+z)/n), i.e., the average of the logs never exceeds the log of the average.  Geometrically, this is due to the fact that the graph of the logarithm function is concave downwards which implies that any point (x,y) that lies in the convex hull of the points (a,log(a)),(b,log(b)),...,(z,log(z)) will satisfy y<=log(x). Now ((a+b+...+z)/n,(log(a)+log(b)+...+log(z))/n) lies in the convex hull of the points (a,log(a)),(b,log(b)),...,(z,log(z)), and hence we see that (log(a)+log(b)+...+log(z))/n<=log((a+b+...+z)/n).

Posted by Richard on 2006-10-29 03:40:20