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Geometrically, since a straight line is the shortest path from a point to another: $$\sum\sqrt{x_i^2+y_i^2}\le \sqrt{\left(\sum x_i\right)^2+\left(\sum y_i\right)^2}$$ Where $x_i,y_i$ are positive reals, and equality occurs when $$\frac{x_i}{y_i}=\frac{\sum x_i}{\sum y_i}$$ That is, all the points lie in the line. But ¿How would we prove it algebraically? The fact that we need that one sequence is a multiple of another makes me think in Cauchy-Shwarz, but I have no other interesing ideas about this.

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By squaring both sides we have that the inequality (with the right direction, $\geq$) is equivalent to: $$\sum_{i\neq j}\sqrt{(x_i^2+y_i^2)(x_j^2+y_j^2)}\geq \sum_{i\neq j}(x_i x_j+y_i y_j)$$ that follows from the Cauchy-Schwarz inequality.

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    $\begingroup$ Thank you! I think I wasn't brave enough to dare square it... $\endgroup$
    – chubakueno
    Nov 6, 2014 at 21:45

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