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A name for this inequality?

  1. Feb 18, 2009 #1
    It's a trivial inequality...but I am curious if there is a popular name for it already. I have not been able to find it through conventional online search.

    a12 + a22 + ... + an2 > a1*a2 + a2*a1 + a2*a3 + ... + an*an-1

    Such as:

    a2 + b2 + c2 > ab + ac + bc + ba + bc + ac = 2(ab + bc + ac)

    Which can nicely be envisioned geometrically as sum of squares will always be bigger than corresponding sum of rectangles.
  2. jcsd
  3. Feb 18, 2009 #2


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    It doesn't look like it's true. Let a=b>c. Then a2+b2=2ab, while c2<2(ac+bc).
  4. Feb 18, 2009 #3


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    There are a lot more of the rectangles than there are the squares.... in fact, another easy source of counterexamples is just to make all the numbers the same! e.g. if they're all 1, then the L.H.S. is n, but the R.H.S. is n(n-1)
  5. Feb 18, 2009 #4
    A very similar inequality is called the rearrangement inequality which works when you rearrange two sets of number, not one.
  6. Feb 19, 2009 #5


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    Short tutorial on the subject.

    First to get it right you need to omit half the terms.
    a2 + b2 + c2 >= ab + bc + ac
    There is a general theorem for vector spaces with inner (dot) products, namely
    A.B=|A||B|cosx, where x is the angle between A and B.

    For the case you are interested in:
    A=(a1,a2,a3,...,an) while B=(a2,a3,...,an,a1)
  7. Feb 20, 2009 #6


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    I had a homework assignment in my first year of undergrad asking for us to prove the Cauchy-Riemann-Schwartz inequality (a.k.a. the triangle inequality), and Googling one or both of these terms seem to generate at least a few results that have a similar form.
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