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## Main Question or Discussion Point

$$\sum_{i=1}^{n} x_i^2 > \frac{1}{n^2}(\sum_{i=1}^{n} x_i)^2$$

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.

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.