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Sum of cross terms vs sum of squares? [on hold]


$fleft(sum X_iright) leq sum f(X_i)$, where $X_igt 0$; for what functions is this true?Maximizing the sum of two numbers, the sum of whose squares is constantComparing sums of surds without any aidsReduce $(a+b)^2(c+d)^2-16abcd$ to sum of squaresDoes a sum of squares become smaller as the number of terms increases?relationship between sum of squares and sumCan we equate coefficients of power series over inequalities?Is $2^n$ always dominate $n^k$, where $k$ is some fixed natural numberFour squares, all of different areas, are cut from a rectangle, leaving a smaller rectangle…Chebyshev's Sum Inequality Proof













1












$begingroup$


What do we know about sum of squares vs sum of cross terms? Does one always dominate the other? Any theorems on that?



e.g
for



$a^2 + b^2 + c^2 < ? > ab + ac + bc $



for any number of terms.



Thank you






inequality rearrangement-inequality







share|cite|improve this question











$endgroup$



put on hold as off-topic by Saad, Cesareo, GNUSupporter 8964民主女神 地下教會, Paul Frost, Eevee Trainer 1 hour ago


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Cesareo, GNUSupporter 8964民主女神 地下教會, Paul Frost, Eevee Trainer
If this question can be reworded to fit the rules in the help center, please edit the question.




















    1












    $begingroup$


    What do we know about sum of squares vs sum of cross terms? Does one always dominate the other? Any theorems on that?



    e.g
    for



    $a^2 + b^2 + c^2 < ? > ab + ac + bc $



    for any number of terms.



    Thank you






    inequality rearrangement-inequality







    share|cite|improve this question











    $endgroup$



    put on hold as off-topic by Saad, Cesareo, GNUSupporter 8964民主女神 地下教會, Paul Frost, Eevee Trainer 1 hour ago


    This question appears to be off-topic. The users who voted to close gave this specific reason:


    • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Cesareo, GNUSupporter 8964民主女神 地下教會, Paul Frost, Eevee Trainer
    If this question can be reworded to fit the rules in the help center, please edit the question.


















      1












      1








      1





      $begingroup$


      What do we know about sum of squares vs sum of cross terms? Does one always dominate the other? Any theorems on that?



      e.g
      for



      $a^2 + b^2 + c^2 < ? > ab + ac + bc $



      for any number of terms.



      Thank you






      inequality rearrangement-inequality







      share|cite|improve this question











      $endgroup$




      What do we know about sum of squares vs sum of cross terms? Does one always dominate the other? Any theorems on that?



      e.g
      for



      $a^2 + b^2 + c^2 < ? > ab + ac + bc $



      for any number of terms.



      Thank you






      inequality rearrangement-inequality




      inequality rearrangement-inequality






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 7 hours ago









      GNUSupporter 8964民主女神 地下教會

      14k82650




      14k82650










      asked 20 hours ago









      YohanRothYohanRoth

      6451715




      6451715




      put on hold as off-topic by Saad, Cesareo, GNUSupporter 8964民主女神 地下教會, Paul Frost, Eevee Trainer 1 hour ago


      This question appears to be off-topic. The users who voted to close gave this specific reason:


      • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Cesareo, GNUSupporter 8964民主女神 地下教會, Paul Frost, Eevee Trainer
      If this question can be reworded to fit the rules in the help center, please edit the question.







      put on hold as off-topic by Saad, Cesareo, GNUSupporter 8964民主女神 地下教會, Paul Frost, Eevee Trainer 1 hour ago


      This question appears to be off-topic. The users who voted to close gave this specific reason:


      • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Cesareo, GNUSupporter 8964民主女神 地下教會, Paul Frost, Eevee Trainer
      If this question can be reworded to fit the rules in the help center, please edit the question.




















          3 Answers
          3






          active

          oldest

          votes


















          1












          $begingroup$

          $$sum_cyc(a^2-ab)=frac12sum_cyc(2a^2-2ab)=frac12sum_cyc(a^2-2ab+b^2)=frac12sum_cyc(a-b)^2geq0.$$
          For $n$ variables there are two versions.





          1. $$a_1^2+a_2^2+...+a_n^2geq a_1a_2+a_2a_3+...+a_na_1.$$



          Proof:
          $$sum_k=1^n(a_k^2-a_ka_k+1)=frac12sum_k=1^n(a_k^2-2a_ka_k+1+a_k+1^2)=frac12sum_k=1^n(a_k-a_k+1)^2geq0.$$
          Here $a_n+1=a_1$.




          1. $$(n-1)(a_1^2+a_2^2+...+a_n^2)geq2(a_1a_2+a_1a_3+...+a_1a_n+a_2a_3+...+a_n-1a_n).$$



          Proof:



          We need to prove that:
          $$(n-1)sum_k=1^na_k^2geq2sum_1leq k<mleq na_ka_m$$ or
          $$sum_1leq k<mleq n(a_k-a_m)^2geq0.$$






          share|cite|improve this answer











          $endgroup$




















            3












            $begingroup$

            for $a,b,cin R(aneq bneq c)$



            $(a-b)^2+(b-c)^2+(c-a)^2>0$



            $a^2+b^2+c^2>ab+bc+ca$






            share|cite|improve this answer









            $endgroup$




















              3












              $begingroup$

              In general, let
              $$f_n(x_1,ldots,x_n)=sum_i,j=1atop i< j^nx_ix_j
              =frac12sum_i,j=1atop ine j^nx_ix_j.$$
              For each $ine j$, $x_ix_jlefrac12(x_i^2+x_j^2)$ (AM/GM)
              with equality iff $x_i=x_j$. Thus
              $$f_n(x_1,ldots,x_n)lefracn-12sum_i=1^nx_i^2$$
              with equality iff $x_1=x_2=cdots=x_n$. From this argument,
              the constant $frac12(n-1)$ is best possible.






              share|cite|improve this answer









              $endgroup$



















                3 Answers
                3






                active

                oldest

                votes








                3 Answers
                3






                active

                oldest

                votes









                active

                oldest

                votes






                active

                oldest

                votes









                1












                $begingroup$

                $$sum_cyc(a^2-ab)=frac12sum_cyc(2a^2-2ab)=frac12sum_cyc(a^2-2ab+b^2)=frac12sum_cyc(a-b)^2geq0.$$
                For $n$ variables there are two versions.





                1. $$a_1^2+a_2^2+...+a_n^2geq a_1a_2+a_2a_3+...+a_na_1.$$



                Proof:
                $$sum_k=1^n(a_k^2-a_ka_k+1)=frac12sum_k=1^n(a_k^2-2a_ka_k+1+a_k+1^2)=frac12sum_k=1^n(a_k-a_k+1)^2geq0.$$
                Here $a_n+1=a_1$.




                1. $$(n-1)(a_1^2+a_2^2+...+a_n^2)geq2(a_1a_2+a_1a_3+...+a_1a_n+a_2a_3+...+a_n-1a_n).$$



                Proof:



                We need to prove that:
                $$(n-1)sum_k=1^na_k^2geq2sum_1leq k<mleq na_ka_m$$ or
                $$sum_1leq k<mleq n(a_k-a_m)^2geq0.$$






                share|cite|improve this answer











                $endgroup$

















                  1












                  $begingroup$

                  $$sum_cyc(a^2-ab)=frac12sum_cyc(2a^2-2ab)=frac12sum_cyc(a^2-2ab+b^2)=frac12sum_cyc(a-b)^2geq0.$$
                  For $n$ variables there are two versions.





                  1. $$a_1^2+a_2^2+...+a_n^2geq a_1a_2+a_2a_3+...+a_na_1.$$



                  Proof:
                  $$sum_k=1^n(a_k^2-a_ka_k+1)=frac12sum_k=1^n(a_k^2-2a_ka_k+1+a_k+1^2)=frac12sum_k=1^n(a_k-a_k+1)^2geq0.$$
                  Here $a_n+1=a_1$.




                  1. $$(n-1)(a_1^2+a_2^2+...+a_n^2)geq2(a_1a_2+a_1a_3+...+a_1a_n+a_2a_3+...+a_n-1a_n).$$



                  Proof:



                  We need to prove that:
                  $$(n-1)sum_k=1^na_k^2geq2sum_1leq k<mleq na_ka_m$$ or
                  $$sum_1leq k<mleq n(a_k-a_m)^2geq0.$$






                  share|cite|improve this answer











                  $endgroup$















                    1












                    1








                    1





                    $begingroup$

                    $$sum_cyc(a^2-ab)=frac12sum_cyc(2a^2-2ab)=frac12sum_cyc(a^2-2ab+b^2)=frac12sum_cyc(a-b)^2geq0.$$
                    For $n$ variables there are two versions.





                    1. $$a_1^2+a_2^2+...+a_n^2geq a_1a_2+a_2a_3+...+a_na_1.$$



                    Proof:
                    $$sum_k=1^n(a_k^2-a_ka_k+1)=frac12sum_k=1^n(a_k^2-2a_ka_k+1+a_k+1^2)=frac12sum_k=1^n(a_k-a_k+1)^2geq0.$$
                    Here $a_n+1=a_1$.




                    1. $$(n-1)(a_1^2+a_2^2+...+a_n^2)geq2(a_1a_2+a_1a_3+...+a_1a_n+a_2a_3+...+a_n-1a_n).$$



                    Proof:



                    We need to prove that:
                    $$(n-1)sum_k=1^na_k^2geq2sum_1leq k<mleq na_ka_m$$ or
                    $$sum_1leq k<mleq n(a_k-a_m)^2geq0.$$






                    share|cite|improve this answer











                    $endgroup$



                    $$sum_cyc(a^2-ab)=frac12sum_cyc(2a^2-2ab)=frac12sum_cyc(a^2-2ab+b^2)=frac12sum_cyc(a-b)^2geq0.$$
                    For $n$ variables there are two versions.





                    1. $$a_1^2+a_2^2+...+a_n^2geq a_1a_2+a_2a_3+...+a_na_1.$$



                    Proof:
                    $$sum_k=1^n(a_k^2-a_ka_k+1)=frac12sum_k=1^n(a_k^2-2a_ka_k+1+a_k+1^2)=frac12sum_k=1^n(a_k-a_k+1)^2geq0.$$
                    Here $a_n+1=a_1$.




                    1. $$(n-1)(a_1^2+a_2^2+...+a_n^2)geq2(a_1a_2+a_1a_3+...+a_1a_n+a_2a_3+...+a_n-1a_n).$$



                    Proof:



                    We need to prove that:
                    $$(n-1)sum_k=1^na_k^2geq2sum_1leq k<mleq na_ka_m$$ or
                    $$sum_1leq k<mleq n(a_k-a_m)^2geq0.$$







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited 19 hours ago

























                    answered 19 hours ago









                    Michael RozenbergMichael Rozenberg

                    107k1894199




                    107k1894199





















                        3












                        $begingroup$

                        for $a,b,cin R(aneq bneq c)$



                        $(a-b)^2+(b-c)^2+(c-a)^2>0$



                        $a^2+b^2+c^2>ab+bc+ca$






                        share|cite|improve this answer









                        $endgroup$

















                          3












                          $begingroup$

                          for $a,b,cin R(aneq bneq c)$



                          $(a-b)^2+(b-c)^2+(c-a)^2>0$



                          $a^2+b^2+c^2>ab+bc+ca$






                          share|cite|improve this answer









                          $endgroup$















                            3












                            3








                            3





                            $begingroup$

                            for $a,b,cin R(aneq bneq c)$



                            $(a-b)^2+(b-c)^2+(c-a)^2>0$



                            $a^2+b^2+c^2>ab+bc+ca$






                            share|cite|improve this answer









                            $endgroup$



                            for $a,b,cin R(aneq bneq c)$



                            $(a-b)^2+(b-c)^2+(c-a)^2>0$



                            $a^2+b^2+c^2>ab+bc+ca$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered 20 hours ago









                            jackyjacky

                            981615




                            981615





















                                3












                                $begingroup$

                                In general, let
                                $$f_n(x_1,ldots,x_n)=sum_i,j=1atop i< j^nx_ix_j
                                =frac12sum_i,j=1atop ine j^nx_ix_j.$$
                                For each $ine j$, $x_ix_jlefrac12(x_i^2+x_j^2)$ (AM/GM)
                                with equality iff $x_i=x_j$. Thus
                                $$f_n(x_1,ldots,x_n)lefracn-12sum_i=1^nx_i^2$$
                                with equality iff $x_1=x_2=cdots=x_n$. From this argument,
                                the constant $frac12(n-1)$ is best possible.






                                share|cite|improve this answer









                                $endgroup$

















                                  3












                                  $begingroup$

                                  In general, let
                                  $$f_n(x_1,ldots,x_n)=sum_i,j=1atop i< j^nx_ix_j
                                  =frac12sum_i,j=1atop ine j^nx_ix_j.$$
                                  For each $ine j$, $x_ix_jlefrac12(x_i^2+x_j^2)$ (AM/GM)
                                  with equality iff $x_i=x_j$. Thus
                                  $$f_n(x_1,ldots,x_n)lefracn-12sum_i=1^nx_i^2$$
                                  with equality iff $x_1=x_2=cdots=x_n$. From this argument,
                                  the constant $frac12(n-1)$ is best possible.






                                  share|cite|improve this answer









                                  $endgroup$















                                    3












                                    3








                                    3





                                    $begingroup$

                                    In general, let
                                    $$f_n(x_1,ldots,x_n)=sum_i,j=1atop i< j^nx_ix_j
                                    =frac12sum_i,j=1atop ine j^nx_ix_j.$$
                                    For each $ine j$, $x_ix_jlefrac12(x_i^2+x_j^2)$ (AM/GM)
                                    with equality iff $x_i=x_j$. Thus
                                    $$f_n(x_1,ldots,x_n)lefracn-12sum_i=1^nx_i^2$$
                                    with equality iff $x_1=x_2=cdots=x_n$. From this argument,
                                    the constant $frac12(n-1)$ is best possible.






                                    share|cite|improve this answer









                                    $endgroup$



                                    In general, let
                                    $$f_n(x_1,ldots,x_n)=sum_i,j=1atop i< j^nx_ix_j
                                    =frac12sum_i,j=1atop ine j^nx_ix_j.$$
                                    For each $ine j$, $x_ix_jlefrac12(x_i^2+x_j^2)$ (AM/GM)
                                    with equality iff $x_i=x_j$. Thus
                                    $$f_n(x_1,ldots,x_n)lefracn-12sum_i=1^nx_i^2$$
                                    with equality iff $x_1=x_2=cdots=x_n$. From this argument,
                                    the constant $frac12(n-1)$ is best possible.







                                    share|cite|improve this answer












                                    share|cite|improve this answer



                                    share|cite|improve this answer










                                    answered 19 hours ago









                                    Lord Shark the UnknownLord Shark the Unknown

                                    105k1161133




                                    105k1161133













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