perplexus.info :: Just Math : Harmonic mean

Harmonic mean (Posted on 2006-10-25)

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.

Submitted by Tristan

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Solution: (Hide)

The harmonic mean is equal to 1/((1/a+1/b+1/c+...)/n).
(1/a+1/b+1/c+...)/n >= (abc...)^(-1/n), since the geometric mean is always less than or equal to the arithmetic mean.
Therefore, 1/((1/a+1/b+1/c+...)/n) <= 1/(abc...)^(-1/n) = (abc...)^(1/n). This proves that the harmonic mean is always less than or equal to the geometric mean.
The proof that the geometric mean is always less than or equal to the arithmetic mean--called the AM/GM inequality--is not a simple one. One such proof can be seen here. Richard also provided some geometric reasoning here.

Comments: ( You must be logged in to post comments.)

Subject	Author	Date
re: Seeing Why GM <= AM	JLo	2006-10-29 10:34:34
Seeing Why GM <= AM	Richard	2006-10-29 03:40:20
Inverse solution	vswitchs	2006-10-25 13:47:46
No need to think	e.g.	2006-10-25 12:25:37

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