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How do I prove this? (summation problem)

  1. Mar 24, 2014 #1
    $$\sum_{i=1}^{n} x_i^2 > \frac{1}{n^2}(\sum_{i=1}^{n} x_i)^2$$

    Note: each x_i is any observation (or statistic) it can be any real number and need not be constrained in anyway whatsoever, though you can take n > 1 and integer (i.e. there is at least two observations and the number of observations is discrete).

    I'm not sure if this true or not, but part of my analysis to a particular problem assumed this was true, and I'm trying to prove it is indeed true (it seems to be case for any examples I come up with).

    So far I came up with,
    $$n^2 \sum_{i=1}^{n} x_i^2 > \sum_{i=1}^{n} x_i^2 + 2\sum_{i \neq j, i > j} x_ix_j$$
    $$(n^2 - 1)\sum_{i=1}^{n}x_i^2 > 2\sum_{i \neq j,\: i > j} x_ix_j$$

    and I'm not sure how to proceed from there.
     
  2. jcsd
  3. Mar 24, 2014 #2

    micromass

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    Are you familiar with the Cauchy-Schwartz inequality?
     
  4. Mar 24, 2014 #3
    Yes I am, but I'm not sure how to use it here. If I was interested in both a x_i and y_i then I would see how to use it here, but here I'm only looking at a x_i.
     
  5. Mar 24, 2014 #4

    micromass

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    Maybe take all ##y_i = 1##?
     
  6. Mar 24, 2014 #5
    Hmm. Alright then by Cauchy-Schwartz I can say,

    $$(\sum_{i=1}^{n} x_i \times 1)^2 \le (\sum_{i=1}^{n}x_i^2) (\sum_{i=1}^{n}1) = n\sum_{i=1}^{n}x_i^2 < n^2 \sum_{i=1}^{n}x_i^2$$

    Which was what I wanted.

    Thanks!
     
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