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# Averages part III

## Question

BMAD    62

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.

## 1 answer to this question

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Rainman    14

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 by Rainman

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