Inequality properties between their multiplications [on hold]











up vote
0
down vote

favorite












If $A>B>C>D$ then will $$(A/B * A/C * A/D) > (B/A * B/C * B/D) > (C/A * C/B * C/D) > (D/A * D/B * D/C)$$ be always true? If not, in what intervals will it not be true?



Obviously $A,B,C,D geq 0$ and they take values in the interval of ${Bbb R}_+$, that is positive real numbers only.






inequality universal-property







share|cite|improve this question









New contributor




mathamateur621 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











put on hold as unclear what you're asking by amWhy, Mark, Zvi, Rebellos, Shailesh 15 hours ago


Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.



















    up vote
    0
    down vote

    favorite












    If $A>B>C>D$ then will $$(A/B * A/C * A/D) > (B/A * B/C * B/D) > (C/A * C/B * C/D) > (D/A * D/B * D/C)$$ be always true? If not, in what intervals will it not be true?



    Obviously $A,B,C,D geq 0$ and they take values in the interval of ${Bbb R}_+$, that is positive real numbers only.






    inequality universal-property







    share|cite|improve this question









    New contributor




    mathamateur621 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.











    put on hold as unclear what you're asking by amWhy, Mark, Zvi, Rebellos, Shailesh 15 hours ago


    Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

















      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      If $A>B>C>D$ then will $$(A/B * A/C * A/D) > (B/A * B/C * B/D) > (C/A * C/B * C/D) > (D/A * D/B * D/C)$$ be always true? If not, in what intervals will it not be true?



      Obviously $A,B,C,D geq 0$ and they take values in the interval of ${Bbb R}_+$, that is positive real numbers only.






      inequality universal-property







      share|cite|improve this question









      New contributor




      mathamateur621 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      If $A>B>C>D$ then will $$(A/B * A/C * A/D) > (B/A * B/C * B/D) > (C/A * C/B * C/D) > (D/A * D/B * D/C)$$ be always true? If not, in what intervals will it not be true?



      Obviously $A,B,C,D geq 0$ and they take values in the interval of ${Bbb R}_+$, that is positive real numbers only.






      inequality universal-property




      inequality universal-property






      share|cite|improve this question









      New contributor




      mathamateur621 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      share|cite|improve this question









      New contributor




      mathamateur621 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.









      share|cite|improve this question




      share|cite|improve this question








      edited yesterday









      Wuestenfux

      2,3541410




      2,3541410






      New contributor




      mathamateur621 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.









      asked yesterday









      mathamateur621

      1




      1




      New contributor




      mathamateur621 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.





      New contributor





      mathamateur621 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






      mathamateur621 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.




      put on hold as unclear what you're asking by amWhy, Mark, Zvi, Rebellos, Shailesh 15 hours ago


      Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.






      put on hold as unclear what you're asking by amWhy, Mark, Zvi, Rebellos, Shailesh 15 hours ago


      Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
























          2 Answers
          2






          active

          oldest

          votes

















          up vote
          1
          down vote













          First of all, we're dividing by $A,B,C,D$ so none can be $0$. Hence, in this answer, we assume $A,B,C,D>0$.



          Anyway, since $A>B>C>D>0$, we know that

          $$A^4>B^4>C^4>D^4$$



          and now since $ABCD>0$, we may divide by $ABCD$ without flipping any signs:



          $$frac{A^3}{BCD}>frac{B^3}{ACD}>frac{C^3}{ABD}>frac{D^3}{ABC}$$



          which is the exact same as the thing you were trying to prove.






          share|cite|improve this answer




























            up vote
            0
            down vote













            Yes---compare term by term e.g. $A/B > B/A$, $A/C > B/C$, $A/D > B/D$ which will give you the first result, and so on.






            share|cite|improve this answer




























              2 Answers
              2






              active

              oldest

              votes








              2 Answers
              2






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes








              up vote
              1
              down vote













              First of all, we're dividing by $A,B,C,D$ so none can be $0$. Hence, in this answer, we assume $A,B,C,D>0$.



              Anyway, since $A>B>C>D>0$, we know that

              $$A^4>B^4>C^4>D^4$$



              and now since $ABCD>0$, we may divide by $ABCD$ without flipping any signs:



              $$frac{A^3}{BCD}>frac{B^3}{ACD}>frac{C^3}{ABD}>frac{D^3}{ABC}$$



              which is the exact same as the thing you were trying to prove.






              share|cite|improve this answer

























                up vote
                1
                down vote













                First of all, we're dividing by $A,B,C,D$ so none can be $0$. Hence, in this answer, we assume $A,B,C,D>0$.



                Anyway, since $A>B>C>D>0$, we know that

                $$A^4>B^4>C^4>D^4$$



                and now since $ABCD>0$, we may divide by $ABCD$ without flipping any signs:



                $$frac{A^3}{BCD}>frac{B^3}{ACD}>frac{C^3}{ABD}>frac{D^3}{ABC}$$



                which is the exact same as the thing you were trying to prove.






                share|cite|improve this answer























                  up vote
                  1
                  down vote










                  up vote
                  1
                  down vote









                  First of all, we're dividing by $A,B,C,D$ so none can be $0$. Hence, in this answer, we assume $A,B,C,D>0$.



                  Anyway, since $A>B>C>D>0$, we know that

                  $$A^4>B^4>C^4>D^4$$



                  and now since $ABCD>0$, we may divide by $ABCD$ without flipping any signs:



                  $$frac{A^3}{BCD}>frac{B^3}{ACD}>frac{C^3}{ABD}>frac{D^3}{ABC}$$



                  which is the exact same as the thing you were trying to prove.






                  share|cite|improve this answer












                  First of all, we're dividing by $A,B,C,D$ so none can be $0$. Hence, in this answer, we assume $A,B,C,D>0$.



                  Anyway, since $A>B>C>D>0$, we know that

                  $$A^4>B^4>C^4>D^4$$



                  and now since $ABCD>0$, we may divide by $ABCD$ without flipping any signs:



                  $$frac{A^3}{BCD}>frac{B^3}{ACD}>frac{C^3}{ABD}>frac{D^3}{ABC}$$



                  which is the exact same as the thing you were trying to prove.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered yesterday









                  vrugtehagel

                  10.7k1549




                  10.7k1549






















                      up vote
                      0
                      down vote













                      Yes---compare term by term e.g. $A/B > B/A$, $A/C > B/C$, $A/D > B/D$ which will give you the first result, and so on.






                      share|cite|improve this answer

























                        up vote
                        0
                        down vote













                        Yes---compare term by term e.g. $A/B > B/A$, $A/C > B/C$, $A/D > B/D$ which will give you the first result, and so on.






                        share|cite|improve this answer























                          up vote
                          0
                          down vote










                          up vote
                          0
                          down vote









                          Yes---compare term by term e.g. $A/B > B/A$, $A/C > B/C$, $A/D > B/D$ which will give you the first result, and so on.






                          share|cite|improve this answer












                          Yes---compare term by term e.g. $A/B > B/A$, $A/C > B/C$, $A/D > B/D$ which will give you the first result, and so on.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered yesterday









                          Richard Martin

                          1,3138




                          1,3138















                              Popular posts from this blog

                              Can a sorcerer learn a 5th-level spell early by creating spell slots using the Font of Magic feature?

                              Image
                              28 6 $begingroup$ Per the Font of Magic feature, sorcerer can use Flexible Casting to create 5th-level spell slots at level 7, even though they are not typically available until level 9. You can transform unexpended sorcery points into one spell slot as a bonus action on your turn. The Creating Spell Slots table shows the cost of creating a spell slot of [5th level is 7] If such a sorcerer levels up while still having this 5th-level spell slot, can they choose a 5th-level spell as the spell they gain upon levelling up? The Spellcasting feature states: Additionally, when you gain a level in this class, you can choose one of the sorcerer spells you know and replace it with another spell from the sorcerer spell list, which also must be of a level for which you have spell slots.
                              Read more

                              Does disintegrating a polymorphed enemy still kill it after the 2018 errata?

                              Image
                              up vote 13 down vote favorite 1 After the 2018 errata, the disintegrate spell description now reads: A creature targeted by this spell must make a Dexterity saving throw. On a failed save, the target takes 10d6 + 40 force damage. The target is disintegrated if this damage leaves it with 0 hit points. If you were to polymorph an enemy into a rat and then disintegrate it, would the enemy be disintegrated or would it just return to its original form? I know that for Druids, it’s not an instant kill anymore, but is this the case for polymorph as well? dnd-5e spells polymorph errata share | improve this question edited 18 hours ago
                              Read more

                              A Topological Invariant for $pi_3(U(n))$

                              Image
                              3 3 $begingroup$ Recently, I saw a construction of topological invariant for $pi_3(U(n))$ with $ngeq 2$ : $$ N=frac{1}{24pi^2}int_{S^3} d^3x epsilon^{ijk} Tr[(U^{-1}partial_{x_i}U)(U^{-1}partial_{x_j}U)(U^{-1}partial_{x_k}U)] , $$ where $Uin U(n)$ depends on $boldsymbol{x}=(x_1,x_2,x_3)in S^3$ , $epsilon^{ijk}$ is the Levi-Civita symbol, $i,j,k=1,2,3$ , and the duplicated indexes are summed over. It is claimed that $N$ is an integer, but why? Update 02/02/2019 I think I got an argument for $n=2$ . In this case, $U=e^{i varphi} q$ with $qin SU(2)$ . Due to the trace and the Levi-Civita symbol in $N$ , $varphi$ does not contribute to $N$ . As $Tr[q^{dagger}partial_i q]=0$ and $(q^{dagger}partial_i q)^{dagger}=-q^{dagger}partial_i q$ , $q^{dagger}partial_i q$ in geneeral has the form $q^{dagger}partia
                              Read more