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An inequality about inner product

  1. Nov 6, 2006 #1
    If α,β,γ are vectors in the Euclid space V, please show that
    |α-β||γ|≤|α-γ||β|+|β-γ||α|,where |α|=√(α,α)
    and point out when the equal mark holds.

    Can someone help me out?
     
  2. jcsd
  3. Nov 6, 2006 #2

    acm

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    This question reeks of Triangle Inequality, I'm assuming the relation: Abs(A-B) <= Abs(A) + Abs(B)

    Abs(A-B)Abs(C) <= Abs(A)Abs(C) + Abs(C)Abs(B)
    Abs(A-C)Abs(B) <= Abs(A)Abs(B) + Abs(C)Abs(B)
    Abs(B-C)Abs(A) <= Abs(B)Abs(A) + Abs(C)Abs(A)

    Abs(C)Abs(B) = Abs(A-C)Abs(B) - Abs(A)Abs(B)
    Abs(A)Abs(C) = Abs(B-C)Abs(A) - Abs(B)Abs(A)

    Abs(A-B)Abs(C) <= Abs(A-C)Abs(B) + Abs(B-C)Abs(A) - 2Abs(A)Abs(B)

    I'm stumped about how to remove the -2Abs(A)Abs(B). :mad:

    Equality if A=B=C
     
  4. Nov 6, 2006 #3
    Thanks very much!

    Note that 2Abs(A)Abs(B)>=0, so
    Abs(A-C)Abs(B) + Abs(B-C)Abs(A) - 2Abs(A)Abs(B) <= Abs(A-C)Abs(B) + Abs(B-C)Abs(A)
    thus, Abs(A-B)Abs(C) <= Abs(A-C)Abs(B) + Abs(B-C)Abs(A) - 2Abs(A)Abs(B)
    <= Abs(A-C)Abs(B) + Abs(B-C)Abs(A).
    It's obvious that if A=B=C the equal mark holds.
    But I think there are some other cases that satisfy the the equality.
     
    Last edited: Nov 6, 2006
  5. Nov 6, 2006 #4

    Office_Shredder

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    How about we go from...

    Abs(A-B)Abs(C) <= Abs(A-C)Abs(B) + Abs(B-C)Abs(A) - 2Abs(A)Abs(B)

    to basic knowledge about numbers, e.g. 2Abs(A)Abs(B) >= 0, so
    Abs(A-C)Abs(B) + Abs(B-C)Abs(A) - 2Abs(A)Abs(B) <= Abs(A-C)Abs(B) + Abs(B-C)Abs(A) which gives you what you want

    Back to basics :p
     
  6. Nov 7, 2006 #5
    Oh,my god!I found a severe mistake. We can't claim that
    Abs(C)Abs(B) = Abs(A-C)Abs(B) - Abs(A)Abs(B)
    Abs(A)Abs(C) = Abs(B-C)Abs(A) - Abs(B)Abs(A)


    So, I'm very sorry to say that we didn't verify the inequality.
     
  7. Nov 7, 2006 #6

    quasar987

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    The correct relationship is

    Abs(C)Abs(B) >= Abs(A-C)Abs(B) - Abs(A)Abs(B)
    Abs(A)Abs(C) >= Abs(B-C)Abs(A) - Abs(B)Abs(A)

    but ASAICS this leads nowhere. :(
     
    Last edited: Nov 7, 2006
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