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AM-GM and power mean inequality proof

  1. Dec 5, 2004 #1
    Does anyone know where I can find (online?) the proofs for the arithmetic mean-geometric mean and the power mean inequalities? Thanks in advance. :smile:
     
  2. jcsd
  3. Dec 5, 2004 #2

    Gokul43201

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    If you want a proof of AM > GM, it's pretty simple :

    Start with :
    [tex](a-b)^2 > 0, ~ a <> b [/tex]
    and work your way towards
    [tex]=>a + b > 2 \sqrt{ab} [/tex]

    What is the power mean inequality - can you write it down ?
     
    Last edited: Dec 5, 2004
  4. Dec 5, 2004 #3
    Grr not sure how to use Latex. I'll give it a stab.

    I get why [tex]AM \geq GM[/tex] for two variables, but what I don't get is how to prove it still works when you extend it to n variables.

    Power mean inequality:

    [tex]P_a=((x^a_1...x^a_n)/n)^{1/a}[/tex], where [tex]x_1,...,x_n \geq 0[/tex]

    Then if [tex]a>b[/tex], [tex]P_a \geq P_b[/tex]

    I think the power mean inequality thing should be obvious... hmm. I mean, I guess you could just subsitute values right?
     
  5. Dec 5, 2004 #4
    Gokul, your proof works only for two terms on the left hand side. There is a proof based on logic which is widely accepted and works for n positive numbers. In essence, it states that the sum of n positive numbers whose product is a constant say p is maximum when all the numbers are equal. This can be proved by contradiction. ascky try this out. I think you should be able to get hold of a proof in standard texts on number theory or even discrete mathematics sometimes (because the method of proof goes like P implies Q but Q is a contradiction so P must be false and so on).

    The general power inequality states that the weighted arithmetic mean of the m-th powers is greater than the m-th power of the weighted arithmetic mean of n numbers except when m is a positve proper fraction (in which case the sign of the inequality reverses):

    [tex]\frac{\sum_{i=0}^{n}w_{i}(a_{i}^{m})}{\sum_{i=0}^{n}w_{i}}\geq (\frac{\sum_{i=0}^{n}w_{i}a_{i}}{\sum_{i=0}^{n}w_{i}})^{m}[/tex]

    If the weights are each equal to 1, then the inequality reduces to the special power inequality which you wish to prove I think. I'll give you a hint: use binomial theorem and consider how the signs change as the value of m changes in [tex](-\infty, 0]U[1,\infty)[/tex] and [tex](0,1)[/tex]. You might want to consult Hall and Knight for more info.

    Hope that helps...

    Cheers
    Vivek
     
    Last edited: Dec 5, 2004
  6. Dec 5, 2004 #5
    Are you sure you want to prove that Pa > Pb? I mean if there's a continued product in the numerator with exponents a and finally the a-th root of it taken, all the a's seem to cancel don't they? Are you sure this is what you meant? For then P_a would be a function of x_{i} and n only for 1<=x<=n (and not a function of a).
     
  7. Dec 5, 2004 #6
    Erp. You're absolutely right... forget that.

    I get how to prove AM-GM now :smile:. I'll have to try that general power inequality later. Thanks for the hints, Vivek.
     
  8. Dec 5, 2004 #7

    Gokul43201

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    maverick,

    I wasn't expecting a more than two-term form of the inequality at the high school level. Guess I misjudged that.
     
  9. Dec 6, 2004 #8
    Well isn't it obvious...the power inequality he has mentioned includes n terms and not 2. But still the starting point has to be a proof for n = 2 even if induction is resorted to. At least n terms is what we do nowadays. ascky, go to mathworld.wolfram.com I don't think they have proofs there but the references are good and google it. I'm not sure you'll find the proof out of the box right away because this isn't covered in standard highschool texts apparently. Thats why I referred you to Hall and Knight Higher Algebra in my earlier post.

    Cheers
    Vivek
     
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