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Few suggestions about cauchy inequality

  1. Feb 18, 2012 #1
    As I can see from the formula of cauchy inequality:
    (a1^2+a2^2+...+an^2)^1/2 . (b1^2+b2^2+...+bn)^1/2 >= a1b1+a2b2 + .... + anbn

    Can I conclude from the above formula that:
    (a1+a2+...+an)^1/2 . (b1+b2+...+bn)^1/2 >= (a1b1)^1/2 + (a2b2)^1/2 +...+ (anbn)^1/2
    by setting a1,...,an = p1^2,...pn^2 and b1,...,bn = q1^2,...qn^2

    Also in my lectures they mentioned this formula
    (a1+a2...+an)(b1+b2+...bn) <= n(a1b1+a2b2+...+anbn)
    but they didnt give a proof for it. Anyone can find me proof for it as I dont know how to prove it.
     
  2. jcsd
  3. Feb 18, 2012 #2
    True, provided the a's and b's are non-negative.

    False. Consider, for example, the case n=2, a1 = b2 = 0, a2 = b1 = 1.
     
  4. Feb 19, 2012 #3
    Your right. Rereading the lecture, they mentioned that that a1,..,an and b1,...,bn are in descending order and they are all positive.
     
    Last edited: Feb 19, 2012
  5. Feb 19, 2012 #4
    How to prove it?
     
  6. Feb 19, 2012 #5
    Have you tried induction?
     
  7. Feb 19, 2012 #6
    The last in inequality is named after Chebyshev. Lots of different spellings, but im sure you can google it. The sequences must be sorted the way, and it is not requiered that the terms are positive.
     
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