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Math Help - Need help on a proof

  1. #1
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    Need help on a proof

    IF a,b >0
    Then prove: AM -HM<= (b-a)^2/4a
    This is what I did

    AM = (a + b)/2
    HM = 2/(1/a + 1/b)

    So when I subtract AM - HM: I got this

    (a+b)/2 - 2ab/(a+b)

    [(a+b)^2 - 4ab]/[2(a+b)]

    which reduces to [(a-b)^2] / [2(a+b)]

    Now what do I do?
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  2. #2
    Super Member malaygoel's Avatar
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    Quote Originally Posted by Nichelle14
    IF a,b >0
    Then prove: AM -HM<= (b-a)^2/4a
    This is what I did

    AM = (a + b)/2
    HM = 2/(1/a + 1/b)

    So when I subtract AM - HM: I got this

    (a+b)/2 - 2ab/(a+b)

    [(a+b)^2 - 4ab]/[2(a+b)]

    which reduces to [(a-b)^2] / [2(a+b)]

    Now what do I do?
    Is there any relation between a and b?
    if a< b
    2a< a+b
    1/2a> 1/(a+b)
    and your proof is complete
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  3. #3
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    I believe you are asking for this realationship a and b are positive.

    I don't understand how your part of the proof makes it complete?

    In other words, how does your part be <= (b-a)^2/4b

    Can you explain some more?
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  4. #4
    Super Member malaygoel's Avatar
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    Quote Originally Posted by Nichelle14
    I believe you are asking for this realationship a and b are positive.

    I don't understand how your part of the proof makes it complete?

    In other words, how does your part be <= (b-a)^2/4b

    Can you explain some more?
    originally you have posted (b-a)^2/4a
    continuing my work
    1/2a > 1/(a+b)
    multiplying both sides by [(b-a)^2]/2 (which is a positive quantity)
    we get
    (b-a)^2/4a >(b-a)^2/2(a+b) RHS is equal to AM-HM
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  5. #5
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    earlier you posted

    if a < b, then 2a < a + b

    Why did you chose to do this?
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  6. #6
    Super Member malaygoel's Avatar
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    Quote Originally Posted by Nichelle14
    earlier you posted

    if a < b, then 2a < a + b

    Why did you chose to do this?
    a and b are the two consecutive terms of a series and they should not be equal
    I had the choice whether a would be smaller or bigger than b because had not mentioned anything.
    I chose a<b because it impies the equation you have to prove.
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  7. #7
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    It is not true.
    What about a=b=2
    -----

    I believe I found a solution.
    As my top sentence says, your inequality is false.

    It is however true when, for x,y>0.
    \phi y\geq  x
    Where,
    \phi is the solution to the cubic,
    2x^3+x^2+3x-1=0.
    Last edited by ThePerfectHacker; June 6th 2006 at 07:55 PM.
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  8. #8
    Super Member malaygoel's Avatar
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    Quote Originally Posted by ThePerfectHacker
    It is not true.
    What about a=b=2
    -----

    I believe I found a solution.
    As my top sentence says, your inequality is false.

    It is however true when, for x,y>0.
    \phi y\geq  x
    Where,
    \phi is the solution to the cubic,
    2x^3+x^2+3x-1=0.
    I don,t think my inequality is false. If you see what we set to prove it includes >= sign. It can be broken into > and = separately. If a=b= any positive real number, AM=GM=HM=a=b which is not of much importance. The real thing lies in inequality part. Hence we have to use a relation between the two terms. However, I don't understand your solution, please explain it.
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