How to prove $frac{(x+y)(y+z)(z+x)}{4xyz}≥frac{x+z}{y+z}+frac{y+z}{x+z}$ for $x,y,z>0$?












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$begingroup$



Prove that for $x,y,z$ positive numbers:
$$
frac{(x+y)(y+z)(z+x)}{4xyz}≥frac{x+z}{y+z}+frac{y+z}{x+z}
$$




I tried to apply MA-MG inequality:
$x+y≥2sqrt{xy}$ and the others and multiply them but it becomes
$frac{(x+y)(y+z)(z+x)}{4xyz}≥2$ but $frac{x+z}{y+z}+frac{y+z}{x+z}≥2$ so it doesn't work like that.
I don't know what to apply here.






inequality cauchy-schwarz-inequality







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    2












    $begingroup$



    Prove that for $x,y,z$ positive numbers:
    $$
    frac{(x+y)(y+z)(z+x)}{4xyz}≥frac{x+z}{y+z}+frac{y+z}{x+z}
    $$




    I tried to apply MA-MG inequality:
    $x+y≥2sqrt{xy}$ and the others and multiply them but it becomes
    $frac{(x+y)(y+z)(z+x)}{4xyz}≥2$ but $frac{x+z}{y+z}+frac{y+z}{x+z}≥2$ so it doesn't work like that.
    I don't know what to apply here.






    inequality cauchy-schwarz-inequality







    share|cite|improve this question











    $endgroup$















      2












      2








      2


      1



      $begingroup$



      Prove that for $x,y,z$ positive numbers:
      $$
      frac{(x+y)(y+z)(z+x)}{4xyz}≥frac{x+z}{y+z}+frac{y+z}{x+z}
      $$




      I tried to apply MA-MG inequality:
      $x+y≥2sqrt{xy}$ and the others and multiply them but it becomes
      $frac{(x+y)(y+z)(z+x)}{4xyz}≥2$ but $frac{x+z}{y+z}+frac{y+z}{x+z}≥2$ so it doesn't work like that.
      I don't know what to apply here.






      inequality cauchy-schwarz-inequality







      share|cite|improve this question











      $endgroup$





      Prove that for $x,y,z$ positive numbers:
      $$
      frac{(x+y)(y+z)(z+x)}{4xyz}≥frac{x+z}{y+z}+frac{y+z}{x+z}
      $$




      I tried to apply MA-MG inequality:
      $x+y≥2sqrt{xy}$ and the others and multiply them but it becomes
      $frac{(x+y)(y+z)(z+x)}{4xyz}≥2$ but $frac{x+z}{y+z}+frac{y+z}{x+z}≥2$ so it doesn't work like that.
      I don't know what to apply here.






      inequality cauchy-schwarz-inequality




      inequality cauchy-schwarz-inequality






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Feb 1 at 16:12









      Michael Rozenberg

      110k1896201




      110k1896201










      asked Feb 1 at 15:57









      reducere csreducere cs

      112




      112






















          2 Answers
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          $begingroup$

          We have
          $$
          frac1 {4yz}+frac1 {4xz}ge frac1{(y+z)^2}+frac1{(x+z)^2}.
          $$           Therefore
          $$
          frac{(x+y)}{4xyz}(y+z)(z+x)geleft(frac1{(y+z)^2}+frac1{(x+z)^2}right)(y+z)(z+x),
          $$           which leads to the desired inequality.






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            $begingroup$

            We need to prove that
            $$frac{(x+y)(x+z)(y+z)}{4xyz}-2geqfrac{x+z}{y+z}+frac{y+z}{x+z}-2$$ or
            $$sum_{cyc}frac{(x-y)^2}{xy}geqfrac{4(x-y)^2}{(y+z)(x+z)},$$ which is true by C-S:
            $$sum_{cyc}frac{(x-y)^2}{xy}geqfrac{(x-y+x-z+z-y)^2}{xy+xz+yz}=$$
            $$=frac{4(x-y)^2}{xy+xz+yz}geqfrac{4(x-y)^2}{z^2+xy+xz+yz}=frac{4(x-y)^2}{(y+z)(x+z)}.$$






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              2 Answers
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              2 Answers
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              active

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              active

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              1












              $begingroup$

              We have
              $$
              frac1 {4yz}+frac1 {4xz}ge frac1{(y+z)^2}+frac1{(x+z)^2}.
              $$               Therefore
              $$
              frac{(x+y)}{4xyz}(y+z)(z+x)geleft(frac1{(y+z)^2}+frac1{(x+z)^2}right)(y+z)(z+x),
              $$               which leads to the desired inequality.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                We have
                $$
                frac1 {4yz}+frac1 {4xz}ge frac1{(y+z)^2}+frac1{(x+z)^2}.
                $$                 Therefore
                $$
                frac{(x+y)}{4xyz}(y+z)(z+x)geleft(frac1{(y+z)^2}+frac1{(x+z)^2}right)(y+z)(z+x),
                $$                 which leads to the desired inequality.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  We have
                  $$
                  frac1 {4yz}+frac1 {4xz}ge frac1{(y+z)^2}+frac1{(x+z)^2}.
                  $$                   Therefore
                  $$
                  frac{(x+y)}{4xyz}(y+z)(z+x)geleft(frac1{(y+z)^2}+frac1{(x+z)^2}right)(y+z)(z+x),
                  $$                   which leads to the desired inequality.






                  share|cite|improve this answer









                  $endgroup$



                  We have
                  $$
                  frac1 {4yz}+frac1 {4xz}ge frac1{(y+z)^2}+frac1{(x+z)^2}.
                  $$                   Therefore
                  $$
                  frac{(x+y)}{4xyz}(y+z)(z+x)geleft(frac1{(y+z)^2}+frac1{(x+z)^2}right)(y+z)(z+x),
                  $$                   which leads to the desired inequality.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Feb 1 at 16:06









                  SongSong

                  18.6k21651




                  18.6k21651























                      0












                      $begingroup$

                      We need to prove that
                      $$frac{(x+y)(x+z)(y+z)}{4xyz}-2geqfrac{x+z}{y+z}+frac{y+z}{x+z}-2$$ or
                      $$sum_{cyc}frac{(x-y)^2}{xy}geqfrac{4(x-y)^2}{(y+z)(x+z)},$$ which is true by C-S:
                      $$sum_{cyc}frac{(x-y)^2}{xy}geqfrac{(x-y+x-z+z-y)^2}{xy+xz+yz}=$$
                      $$=frac{4(x-y)^2}{xy+xz+yz}geqfrac{4(x-y)^2}{z^2+xy+xz+yz}=frac{4(x-y)^2}{(y+z)(x+z)}.$$






                      share|cite|improve this answer









                      $endgroup$


















                        0












                        $begingroup$

                        We need to prove that
                        $$frac{(x+y)(x+z)(y+z)}{4xyz}-2geqfrac{x+z}{y+z}+frac{y+z}{x+z}-2$$ or
                        $$sum_{cyc}frac{(x-y)^2}{xy}geqfrac{4(x-y)^2}{(y+z)(x+z)},$$ which is true by C-S:
                        $$sum_{cyc}frac{(x-y)^2}{xy}geqfrac{(x-y+x-z+z-y)^2}{xy+xz+yz}=$$
                        $$=frac{4(x-y)^2}{xy+xz+yz}geqfrac{4(x-y)^2}{z^2+xy+xz+yz}=frac{4(x-y)^2}{(y+z)(x+z)}.$$






                        share|cite|improve this answer









                        $endgroup$
















                          0












                          0








                          0





                          $begingroup$

                          We need to prove that
                          $$frac{(x+y)(x+z)(y+z)}{4xyz}-2geqfrac{x+z}{y+z}+frac{y+z}{x+z}-2$$ or
                          $$sum_{cyc}frac{(x-y)^2}{xy}geqfrac{4(x-y)^2}{(y+z)(x+z)},$$ which is true by C-S:
                          $$sum_{cyc}frac{(x-y)^2}{xy}geqfrac{(x-y+x-z+z-y)^2}{xy+xz+yz}=$$
                          $$=frac{4(x-y)^2}{xy+xz+yz}geqfrac{4(x-y)^2}{z^2+xy+xz+yz}=frac{4(x-y)^2}{(y+z)(x+z)}.$$






                          share|cite|improve this answer









                          $endgroup$



                          We need to prove that
                          $$frac{(x+y)(x+z)(y+z)}{4xyz}-2geqfrac{x+z}{y+z}+frac{y+z}{x+z}-2$$ or
                          $$sum_{cyc}frac{(x-y)^2}{xy}geqfrac{4(x-y)^2}{(y+z)(x+z)},$$ which is true by C-S:
                          $$sum_{cyc}frac{(x-y)^2}{xy}geqfrac{(x-y+x-z+z-y)^2}{xy+xz+yz}=$$
                          $$=frac{4(x-y)^2}{xy+xz+yz}geqfrac{4(x-y)^2}{z^2+xy+xz+yz}=frac{4(x-y)^2}{(y+z)(x+z)}.$$







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Feb 1 at 16:06









                          Michael RozenbergMichael Rozenberg

                          110k1896201




                          110k1896201






























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