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How do I prove this? (summation problem)

by Gridvvk
Tags: prove, summation
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Gridvvk
#1
Mar24-14, 07:58 PM
P: 56
$$\sum_{i=1}^{n} x_i^2 > \frac{1}{n^2}(\sum_{i=1}^{n} x_i)^2$$

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.
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micromass
#2
Mar24-14, 08:00 PM
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Are you familiar with the Cauchy-Schwartz inequality?
Gridvvk
#3
Mar24-14, 08:09 PM
P: 56
Quote Quote by micromass View Post
Are you familiar with the Cauchy-Schwartz inequality?
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.

micromass
#4
Mar24-14, 08:09 PM
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How do I prove this? (summation problem)

Quote Quote by Gridvvk View Post
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.
Maybe take all ##y_i = 1##?
Gridvvk
#5
Mar24-14, 08:17 PM
P: 56
Quote Quote by micromass View Post
Maybe take all ##y_i = 1##?
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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