$${A_i} = \left( {\begin{array}{*{20}{c}}{{A_{i1}}}\\{{A_{i2}}}\\ \vdots \\{{A_{in}}}\end{array}} \right),{B_i} = \left( {\begin{array}{*{20}{c}}{{B_{i1}}}\\{{B_{i2}}}\\ \vdots \\{{B_{in}}}\end{array}} \right).{B_{ij}},{A_{ij}} \in R
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prove
$$\sum\limits_{1 \le i,j \le n} {\left| {{A_i} - {B_j}} \right|} \ge \sum\limits_{1 \le i < j \le n} {\left( {\left| {{A_i} - {A_j}} \right| + \left| {{B_i} - {B_j}} \right|} \right)}
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$$
with equality only in
$$\left\{ {{A_i}} \right\} = \left\{ {{B_i}} \right\}
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$$
I encountered this problem while doing an engineering research.
I did lots of tests using computer program, and the inequality stands.
To prove it, I have tried the triangle inequality, mathematical induction and anything else that I can think of, and I failed.
In the triangle inequality, the number of the items in one side is twice as the other side. However,in this inequality there are n^2 items in LHS and n*(n-1) items in RHS.
Post Closed as "Not suitable for this site" by Will Jagy, Andrés E. Caicedo, Boris Bukh, Marco Golla, Alex Degtyarev