Averages part III

[BMAD]

This post will possibly help with "averages part II"

given two positive real numbers a and b, there are four ways of computing the "mean":

am= arithmetic mean = (a+b)/2

gm = geometric mean = sqrt[a b]

hm = harmonic mean = 2 / (1/a + 1/b)

rms = root mean square = sqrt[(a^2+b^2)/2]

Rank these from smallest to largest proving your ranks. If one type is not always less than another, explain when it is equal or when it is more.

[Rainman]

A

² - G² = (a²+2ab+b²)/4 - ab = (a²-2ab+b²)/4 = (a-b)²/4, which is at least 0 with equality only for a=b. Hence A is greater than or equal to G, with equality only for a=b.

Lemma: (a+b)² is greater than or equal to 4ab, with equality only for a=b.

Proof: (a+b)²-4ab = (a-b)², which is greater than or equal to 0, with equality only for a=b.

H = 2/((a+b)/ab) = 2ab/(a+b)

(H/G)² = 4ab/(a+b)², which is less than or equal to 1 by Lemma. Hence H is less than or equal to G, with equality only for a=b.

R² - A² = (a²+b²)/2 - (a²+2ab+b²)/4 = (2a²+2b²)/4 - (a²+2ab+b²)/4 = (a²-2ab+b²)/4 = ((a-b)/2)², which is at least 0 with equality only for a=b.

Conclusion: R>A>G>H, unless a=b in which case they are all equal.

Edit: ranking them from smallest to largest, that would be H, G, A, R

Edited July 30, 2013 by Rainman