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$${A_i} = \left( {\begin{array}{*{20}{c}}{{A_{i1}}}\\{{A_{i2}}}\\ \vdots \\{{A_{in}}}\end{array}} \right),{B_i} = \left( {\begin{array}{*{20}{c}}{{B_{i1}}}\\{{B_{i2}}}\\ \vdots \\{{B_{in}}}\end{array}} \right).{B_{ij}},{A_{ij}} \in R % MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbWexLMBbXgBd9gzLbvyNv2CaeHbl7mZLdGeaGqiVu0Je9sqqr % pepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9vqaqpepm0xbba9pwe9Q8fs % 0-yqaqpepae9pg0FirpepeKkFr0xfr-xfr-xb9adbaqaaeGaciGaai % aabeqaamaabaabauaakeaacaWGbbWaaSbaaSqaaiaadMgaaeqaaOGa % eyypa0ZaaeWaaeaafaqabeabbaaaaeaacaWGbbWaaSbaaSqaaiaadM % gacaaIXaaabeaaaOqaaiaadgeadaWgaaWcbaGaamyAaiaaikdaaeqa % aaGcbaGaeSO7I0eabaGaamyqamaaBaaaleaacaWGPbGaamOBaaqaba % aaaaGccaGLOaGaayzkaaGaaiilaiaadkeadaWgaaWcbaGaamyAaaqa % baGccqGH9aqpdaqadaqaauaabeqaeeaaaaqaaiaadkeadaWgaaWcba % GaamyAaiaaigdaaeqaaaGcbaGaamOqamaaBaaaleaacaWGPbGaaGOm % aaqabaaakeaacqWIUlstaeaacaWGcbWaaSbaaSqaaiaadMgacaWGUb % aabeaaaaaakiaawIcacaGLPaaacaGGUaGaamOqamaaBaaaleaacaWG % PbGaamOAaaqabaGccaGGSaGaamyqamaaBaaaleaacaWGPbGaamOAaa % qabaGccqGHiiIZcaWGsbaaaa!6685! $$ prove $$\sum\limits_{1 \le i,j \le n} {\left| {{A_i} - {B_j}} \right|} \ge \sum\limits_{1 \le i < j \le n} {\left( {\left| {{A_i} - {A_j}} \right| + \left| {{B_i} - {B_j}} \right|} \right)} % MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbWexLMBbXgBd9gzLbvyNv2CaeHbl7mZLdGeaGqiVu0Je9sqqr % pepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9vqaqpepm0xbba9pwe9Q8fs % 0-yqaqpepae9pg0FirpepeKkFr0xfr-xfr-xb9adbaqaaeGaciGaai % aabeqaamaabaabauaakeaadaaeqbqaamaaemaabaGaamyqamaaBaaa % leaacaWGPbaabeaakiabgkHiTiaadkeadaWgaaWcbaGaamOAaaqaba % aakiaawEa7caGLiWoaaSqaaiaaigdacqGHKjYOcaWGPbGaaiilaiaa % dQgacqGHKjYOcaWGUbaabeqdcqGHris5aOGaeyyzIm7aaabuaeaada % qadaqaamaaemaabaGaamyqamaaBaaaleaacaWGPbaabeaakiabgkHi % TiaadgeadaWgaaWcbaGaamOAaaqabaaakiaawEa7caGLiWoacqGHRa % WkdaabdaqaaiaadkeadaWgaaWcbaGaamyAaaqabaGccqGHsislcaWG % cbWaaSbaaSqaaiaadQgaaeqaaaGccaGLhWUaayjcSdaacaGLOaGaay % zkaaaaleaacaaIXaGaeyizImQaamyAaiabgYda8iaadQgacqGHKjYO % caWGUbaabeqdcqGHris5aaaa!6EEF! $$ with equality only in $$\left\{ {{A_i}} \right\} = \left\{ {{B_i}} \right\} % MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbWexLMBbXgBd9gzLbvyNv2CaeHbl7mZLdGeaGqiVu0Je9sqqr % pepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9vqaqpepm0xbba9pwe9Q8fs % 0-yqaqpepae9pg0FirpepeKkFr0xfr-xfr-xb9adbaqaaeGaciGaai % aabeqaamaabaabauaakabaaaaaaaaapeqaa8aadaGadaqaaiaadgea % daWgaaWcbaGaamyAaaqabaaakiaawUhacaGL9baacqGH9aqpdaGada % qaaiaadkeadaWgaaWcbaGaamyAaaqabaaakiaawUhacaGL9baaaaa!488D! $$ 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.

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    $\begingroup$ I cannot see what makes this question off-topic. The inequality is obviously false for an arbitrary metric space, but the restriction that the vectors are in $\mathbb{R}^n$ makes it quite plausible. Can anyone explain why this was put on hold? $\endgroup$ Jul 26, 2015 at 11:24
  • $\begingroup$ @Dongryul, evidently, it was put on hold because 5 people agreed that it was not about research-level mathematics. If you and/or Bob can convince 5 people that it is about research-level mathematics, you can get it reopened. $\endgroup$ Jul 27, 2015 at 6:17

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