Let be an integer and be real numbers satisfying . If then prove that .

The conditions hint at inequalities involving an order, such as the rearrangement and Chebychev inequalities. Also note that for all is an equality case, hence we should try to use inequalities in such a way that matches the equality case.

The RHS can be rewritten as . That is, .

Now Chebychev inequality gives . Cancelling the denominators, we get the desired result.

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