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RMS-AM-GM-HM Inequality

  1. Nov 13, 2013 #1
    1. The problem statement, all variables and given/known data

    I am asked to discover and prove the inequality relationship between root mean square, arithmetic mean, geometric mean, and harmonic mean.

    2. Relevant equations
    Let a,b, be non-negative integers.
    (a-b)2 ≥ 0 and (√a-√b)2 ≥ 0

    3. The attempt at a solution

    Using (a-b)2 ≥ 0 and (√a-√b)2 ≥ 0, I was able to show that AM ≥ GM , GM ≥ HM, and RMS ≥ GM, but I haven't really been able to show that RMS ≥ AM and I was wondering if someone could point me in the right direction.

    I used (a-b)2 ≥ 0 and did some algebra to show that √((a2+b2)/2) ≥ √ab

    But I don't know if I can use that to show RMS ≥ AM.

    Thanks in advance to anyone who can offer some insight.
     
  2. jcsd
  3. Nov 13, 2013 #2

    Dick

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    You know a^2+b^2>=2ab. Here's a hint. Add a^2+b^2 to both sides.
     
  4. Nov 13, 2013 #3
    Thanks for the help! It clarified the step I was missing. From there, I would get (a^2+b^2)/2>=((a+b)^2)/4 and then take the square root to get RMS>=AM.
     
  5. Nov 13, 2013 #4

    Dick

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    You're welcome. Good use of the hint!
     
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