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This is a problem I saw on another website and that didn't have a solution. I also suspect it doesn't have a trivial solution, hence why I post it here: let $\{x_i\}_{i=1}^n$ and $\{y_i\}_{i=1}^n$ be complex. Then, the following inequality holds

$$\sum_{i, j = 1}^n |x_i-y_j|+|x_j-y_i| - |x_i-x_j| - |y_i-y_j| \ge 0$$

I don't see how induction would work here, but it might. One idea I had was to get the sum in the form $\sum (\dots) \frac{x_i-y_i}{|x_i-y_i|}\frac{x_j-y_j}{|x_j-y_j|}$ since the summation terms are antisymmetric under the exchanges $x_i\Leftrightarrow y_i$ and $x_j\Leftrightarrow y_j$. But I couldn't do it even if the x's and y's are real. Any ideas for either the real or complex case?

Ivan
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  • Have you tried to use the triangle inequality and an induction? The case $n=2$ follows easily from $|x_1-x_2| \leq |x_1-y_1|+|y_1-x_2|$ and $|y_1-y_2| \leq |y_1-x_1|-|x_1-y_2|$. – Stefano Apr 07 '20 at 23:04
  • @CalvinLin, yes, it is the same question! Thank you! – Ivan Apr 08 '20 at 01:42
  • @Stefano If you see the linked problem, $n=2$ is easy via triangle inequality, but thereafter it' (seems) "impossible" to extend the argument. – Calvin Lin Apr 08 '20 at 01:44

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