$${A_i} = \left( {\begin{array}{*{20}{c}}{{A_{i1}}}\\{{A_{i2}}}\\ \vdots \\{{A_{in}}}\end{array}} \right),{B_i} = \left( {\begin{array}{*{20}{c}}{{B_{i1}}}\\{{B_{i2}}}\\ \vdots \\{{B_{in}}}\end{array}} \right).{B_{ij}},{A_{ij}} \in R
% MathType!MTEF!2!1!+-
% feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn
% hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr
% 4rNCHbWexLMBbXgBd9gzLbvyNv2CaeHbl7mZLdGeaGqiVu0Je9sqqr
% pepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9vqaqpepm0xbba9pwe9Q8fs
% 0-yqaqpepae9pg0FirpepeKkFr0xfr-xfr-xb9adbaqaaeGaciGaai
% aabeqaamaabaabauaakeaacaWGbbWaaSbaaSqaaiaadMgaaeqaaOGa
% eyypa0ZaaeWaaeaafaqabeabbaaaaeaacaWGbbWaaSbaaSqaaiaadM
% gacaaIXaaabeaaaOqaaiaadgeadaWgaaWcbaGaamyAaiaaikdaaeqa
% aaGcbaGaeSO7I0eabaGaamyqamaaBaaaleaacaWGPbGaamOBaaqaba
% aaaaGccaGLOaGaayzkaaGaaiilaiaadkeadaWgaaWcbaGaamyAaaqa
% baGccqGH9aqpdaqadaqaauaabeqaeeaaaaqaaiaadkeadaWgaaWcba
% GaamyAaiaaigdaaeqaaaGcbaGaamOqamaaBaaaleaacaWGPbGaaGOm
% aaqabaaakeaacqWIUlstaeaacaWGcbWaaSbaaSqaaiaadMgacaWGUb
% aabeaaaaaakiaawIcacaGLPaaacaGGUaGaamOqamaaBaaaleaacaWG
% PbGaamOAaaqabaGccaGGSaGaamyqamaaBaaaleaacaWGPbGaamOAaa
% qabaGccqGHiiIZcaWGsbaaaa!6685!
$$
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.
$\begingroup$
$\endgroup$
2
-
1$\begingroup$ 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? $\endgroup$– Dongryul KimCommented 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. $\endgroup$– Gerry MyersonCommented Jul 27, 2015 at 6:17
Add a comment
|