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Let $(x_i)_{i=1}^{i=n}, (y_i)_{i=1}^{i=n}$ are sequence of positive numbers with $\sum_{i=1}^{n}x_i=\sum_{i=1}^{n}y_i=1$. Prove that $\sum_{i=1}^{n}x_iy_i$ is maximized when $x_i$ and $y_i$ are roughly the same size (More precisely, when $x_1>x_2>\cdots > x_n$ and $y_1>x_y>\cdots > y_n$).

I have seen an algebraic proof long time ago, but I couldn't write the solution again. Can anyone help?

Andy
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  • This may help: https://en.wikipedia.org/wiki/Rearrangement_inequality – Minus One-Twelfth Sep 25 '19 at 10:49
  • What exactly do you mean? If the only constraints are that the $x_i$ are positive with sum 1, and similarly for the $y_i$, then you maximise by taking $x_1=y_1=1-\epsilon$ and $x_i=y_i=\frac{\epsilon}{n-1}$ for $i>1$, which gets as close as you like to 1 for $\epsilon$ sufficiently small. Are you fixing the $x_i$ and asking which $y_i$ maximise, or what? – almagest Sep 25 '19 at 14:09

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