Posted 30 Jul 2013 · Report post 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. 0 Share this post Link to post Share on other sites

0 Posted 30 Jul 2013 (edited) · Report post A^{2} - G^{2} = (a^{2}+2ab+b^{2})/4 - ab = (a^{2}-2ab+b^{2})/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. R^{2} - A^{2} = (a^{2}+b^{2})/2 - (a^{2}+2ab+b^{2})/4 = (2a^{2}+2b^{2})/4 - (a^{2}+2ab+b^{2})/4 = (a^{2}-2ab+b^{2})/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 30 Jul 2013 by Rainman 0 Share this post Link to post Share on other sites

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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.

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