Inequality with $(x+y)(y+z)(z+w)(w+x)=1$












3














Let $x,y,z,w>0$ and such that
$$(x+y)(y+z)(z+w)(w+x)=1.$$
Show that
$$sqrt[3]{xyz}+sqrt[3]{yzw}+sqrt[3]{zwx}+sqrt[3]{wxy}le 2.$$



I'm trying to use Holder's inequality
$$(sqrt[3]{xyz}+sqrt[3]{yzw})^3
le (x+y)(y+z)(z+w)$$
$$(sqrt[3]{zwx}+sqrt[3]{wxy})^3le (z+w)(w+x)(x+y)$$
so
$$sqrt[3]{xyz}+sqrt[3]{yzw}+sqrt[3]{zwx}+sqrt[3]{wxy}le sqrt[3]{(x+y)(y+z)(z+w)}+sqrt[3]{(z+w)(w+x)(x+y)}.$$






inequality radicals a.m.-g.m.-inequality holder-inequality







share|cite|improve this question





























    3














    Let $x,y,z,w>0$ and such that
    $$(x+y)(y+z)(z+w)(w+x)=1.$$
    Show that
    $$sqrt[3]{xyz}+sqrt[3]{yzw}+sqrt[3]{zwx}+sqrt[3]{wxy}le 2.$$



    I'm trying to use Holder's inequality
    $$(sqrt[3]{xyz}+sqrt[3]{yzw})^3
    le (x+y)(y+z)(z+w)$$
    $$(sqrt[3]{zwx}+sqrt[3]{wxy})^3le (z+w)(w+x)(x+y)$$
    so
    $$sqrt[3]{xyz}+sqrt[3]{yzw}+sqrt[3]{zwx}+sqrt[3]{wxy}le sqrt[3]{(x+y)(y+z)(z+w)}+sqrt[3]{(z+w)(w+x)(x+y)}.$$






    inequality radicals a.m.-g.m.-inequality holder-inequality







    share|cite|improve this question



























      3












      3








      3


      3





      Let $x,y,z,w>0$ and such that
      $$(x+y)(y+z)(z+w)(w+x)=1.$$
      Show that
      $$sqrt[3]{xyz}+sqrt[3]{yzw}+sqrt[3]{zwx}+sqrt[3]{wxy}le 2.$$



      I'm trying to use Holder's inequality
      $$(sqrt[3]{xyz}+sqrt[3]{yzw})^3
      le (x+y)(y+z)(z+w)$$
      $$(sqrt[3]{zwx}+sqrt[3]{wxy})^3le (z+w)(w+x)(x+y)$$
      so
      $$sqrt[3]{xyz}+sqrt[3]{yzw}+sqrt[3]{zwx}+sqrt[3]{wxy}le sqrt[3]{(x+y)(y+z)(z+w)}+sqrt[3]{(z+w)(w+x)(x+y)}.$$






      inequality radicals a.m.-g.m.-inequality holder-inequality







      share|cite|improve this question















      Let $x,y,z,w>0$ and such that
      $$(x+y)(y+z)(z+w)(w+x)=1.$$
      Show that
      $$sqrt[3]{xyz}+sqrt[3]{yzw}+sqrt[3]{zwx}+sqrt[3]{wxy}le 2.$$



      I'm trying to use Holder's inequality
      $$(sqrt[3]{xyz}+sqrt[3]{yzw})^3
      le (x+y)(y+z)(z+w)$$
      $$(sqrt[3]{zwx}+sqrt[3]{wxy})^3le (z+w)(w+x)(x+y)$$
      so
      $$sqrt[3]{xyz}+sqrt[3]{yzw}+sqrt[3]{zwx}+sqrt[3]{wxy}le sqrt[3]{(x+y)(y+z)(z+w)}+sqrt[3]{(z+w)(w+x)(x+y)}.$$






      inequality radicals a.m.-g.m.-inequality holder-inequality




      inequality radicals a.m.-g.m.-inequality holder-inequality






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 29 '18 at 5:23









      Michael Rozenberg

      97.4k1589188




      97.4k1589188










      asked Dec 29 '18 at 1:19









      inequalityinequality

      701520




      701520






















          2 Answers
          2






          active

          oldest

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          3














          Since $$prod_{cyc}(x+y)-sum_{cyc}xsum_{cyc}xyz=(xz-yw)^2geq0,$$ by Holder we obtain:



          $$2=2left(prod_{cyc}(x+y)right)^{frac{1}{4}}geq2left(sum_{cyc}xsum_{cyc}xyzright)^{frac{1}{4}}=$$
          $$=sqrt[4]{left(4^2sum_{cyc}xyzright)sum_{cyc}x}geqsqrt[4]{left(sum_{cyc}sqrt[3]{xyz}right)^3sum_{cyc}x}.$$
          Thus, it's enough to prove that
          $$sum_{cyc}xgeqsum_{cyc}sqrt[3]{xyz},$$ which is true by AM-GM:
          $$sum_{cyc}x=frac{1}{3}sum_{cyc}(x+y+z)geqfrac{1}{3}sum_{cyc}3sqrt[3]{xyz}=sum_{cyc}sqrt[3]{xyz}.$$
          Done!



          I used Holder for two sequences:
          $$16sum_{cyc}xyz=left(sum_{cyc}1right)^2sum_{cyc}xyzgeqleft(sum_{cyc}sqrt[3]{1^2xyz}right)^3=left(sum_{cyc}sqrt[3]{xyz}right)^3$$






          share|cite|improve this answer























          • Can you clarify the Holder part?
            – Pratyush Sarkar
            Dec 29 '18 at 5:51






          • 1




            @Pratyush Sarkar I added something. See now.
            – Michael Rozenberg
            Dec 29 '18 at 6:46






          • 1




            Ah, thanks that helps a lot. Nice concise answer.
            – Pratyush Sarkar
            Dec 29 '18 at 6:56





















          1














          This does not seem the best, but using only Holder's inequality, we can get
          $$begin{eqnarray}
          S^{12}&le& (x+y+x+y)(y+y+x+x)(y+y+x+x)(z+y+z+y)\&&(y+z+z+y)(y+z+z+y)(z+w+z+w)(z+z+w+w)\&&(z+z+w+w)(x+w+x+w)(x+w+w+x)(x+w+w+x) = 2^{12}left(prod_{text{cyc}}(x+y)right)^3 =2^{12},
          end{eqnarray}$$          for $S= sqrt[3]{xyz}+sqrt[3]{yzw}+sqrt[3]{zwx}+sqrt[3]{wxy}.$






          share|cite|improve this answer





















          • Very nice solution! +1
            – Michael Rozenberg
            Dec 30 '18 at 9:53













          Your Answer





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          2 Answers
          2






          active

          oldest

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          2 Answers
          2






          active

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          active

          oldest

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          active

          oldest

          votes









          3














          Since $$prod_{cyc}(x+y)-sum_{cyc}xsum_{cyc}xyz=(xz-yw)^2geq0,$$ by Holder we obtain:



          $$2=2left(prod_{cyc}(x+y)right)^{frac{1}{4}}geq2left(sum_{cyc}xsum_{cyc}xyzright)^{frac{1}{4}}=$$
          $$=sqrt[4]{left(4^2sum_{cyc}xyzright)sum_{cyc}x}geqsqrt[4]{left(sum_{cyc}sqrt[3]{xyz}right)^3sum_{cyc}x}.$$
          Thus, it's enough to prove that
          $$sum_{cyc}xgeqsum_{cyc}sqrt[3]{xyz},$$ which is true by AM-GM:
          $$sum_{cyc}x=frac{1}{3}sum_{cyc}(x+y+z)geqfrac{1}{3}sum_{cyc}3sqrt[3]{xyz}=sum_{cyc}sqrt[3]{xyz}.$$
          Done!



          I used Holder for two sequences:
          $$16sum_{cyc}xyz=left(sum_{cyc}1right)^2sum_{cyc}xyzgeqleft(sum_{cyc}sqrt[3]{1^2xyz}right)^3=left(sum_{cyc}sqrt[3]{xyz}right)^3$$






          share|cite|improve this answer























          • Can you clarify the Holder part?
            – Pratyush Sarkar
            Dec 29 '18 at 5:51






          • 1




            @Pratyush Sarkar I added something. See now.
            – Michael Rozenberg
            Dec 29 '18 at 6:46






          • 1




            Ah, thanks that helps a lot. Nice concise answer.
            – Pratyush Sarkar
            Dec 29 '18 at 6:56


















          3














          Since $$prod_{cyc}(x+y)-sum_{cyc}xsum_{cyc}xyz=(xz-yw)^2geq0,$$ by Holder we obtain:



          $$2=2left(prod_{cyc}(x+y)right)^{frac{1}{4}}geq2left(sum_{cyc}xsum_{cyc}xyzright)^{frac{1}{4}}=$$
          $$=sqrt[4]{left(4^2sum_{cyc}xyzright)sum_{cyc}x}geqsqrt[4]{left(sum_{cyc}sqrt[3]{xyz}right)^3sum_{cyc}x}.$$
          Thus, it's enough to prove that
          $$sum_{cyc}xgeqsum_{cyc}sqrt[3]{xyz},$$ which is true by AM-GM:
          $$sum_{cyc}x=frac{1}{3}sum_{cyc}(x+y+z)geqfrac{1}{3}sum_{cyc}3sqrt[3]{xyz}=sum_{cyc}sqrt[3]{xyz}.$$
          Done!



          I used Holder for two sequences:
          $$16sum_{cyc}xyz=left(sum_{cyc}1right)^2sum_{cyc}xyzgeqleft(sum_{cyc}sqrt[3]{1^2xyz}right)^3=left(sum_{cyc}sqrt[3]{xyz}right)^3$$






          share|cite|improve this answer























          • Can you clarify the Holder part?
            – Pratyush Sarkar
            Dec 29 '18 at 5:51






          • 1




            @Pratyush Sarkar I added something. See now.
            – Michael Rozenberg
            Dec 29 '18 at 6:46






          • 1




            Ah, thanks that helps a lot. Nice concise answer.
            – Pratyush Sarkar
            Dec 29 '18 at 6:56
















          3












          3








          3






          Since $$prod_{cyc}(x+y)-sum_{cyc}xsum_{cyc}xyz=(xz-yw)^2geq0,$$ by Holder we obtain:



          $$2=2left(prod_{cyc}(x+y)right)^{frac{1}{4}}geq2left(sum_{cyc}xsum_{cyc}xyzright)^{frac{1}{4}}=$$
          $$=sqrt[4]{left(4^2sum_{cyc}xyzright)sum_{cyc}x}geqsqrt[4]{left(sum_{cyc}sqrt[3]{xyz}right)^3sum_{cyc}x}.$$
          Thus, it's enough to prove that
          $$sum_{cyc}xgeqsum_{cyc}sqrt[3]{xyz},$$ which is true by AM-GM:
          $$sum_{cyc}x=frac{1}{3}sum_{cyc}(x+y+z)geqfrac{1}{3}sum_{cyc}3sqrt[3]{xyz}=sum_{cyc}sqrt[3]{xyz}.$$
          Done!



          I used Holder for two sequences:
          $$16sum_{cyc}xyz=left(sum_{cyc}1right)^2sum_{cyc}xyzgeqleft(sum_{cyc}sqrt[3]{1^2xyz}right)^3=left(sum_{cyc}sqrt[3]{xyz}right)^3$$






          share|cite|improve this answer














          Since $$prod_{cyc}(x+y)-sum_{cyc}xsum_{cyc}xyz=(xz-yw)^2geq0,$$ by Holder we obtain:



          $$2=2left(prod_{cyc}(x+y)right)^{frac{1}{4}}geq2left(sum_{cyc}xsum_{cyc}xyzright)^{frac{1}{4}}=$$
          $$=sqrt[4]{left(4^2sum_{cyc}xyzright)sum_{cyc}x}geqsqrt[4]{left(sum_{cyc}sqrt[3]{xyz}right)^3sum_{cyc}x}.$$
          Thus, it's enough to prove that
          $$sum_{cyc}xgeqsum_{cyc}sqrt[3]{xyz},$$ which is true by AM-GM:
          $$sum_{cyc}x=frac{1}{3}sum_{cyc}(x+y+z)geqfrac{1}{3}sum_{cyc}3sqrt[3]{xyz}=sum_{cyc}sqrt[3]{xyz}.$$
          Done!



          I used Holder for two sequences:
          $$16sum_{cyc}xyz=left(sum_{cyc}1right)^2sum_{cyc}xyzgeqleft(sum_{cyc}sqrt[3]{1^2xyz}right)^3=left(sum_{cyc}sqrt[3]{xyz}right)^3$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 29 '18 at 6:45

























          answered Dec 29 '18 at 5:00









          Michael RozenbergMichael Rozenberg

          97.4k1589188




          97.4k1589188












          • Can you clarify the Holder part?
            – Pratyush Sarkar
            Dec 29 '18 at 5:51






          • 1




            @Pratyush Sarkar I added something. See now.
            – Michael Rozenberg
            Dec 29 '18 at 6:46






          • 1




            Ah, thanks that helps a lot. Nice concise answer.
            – Pratyush Sarkar
            Dec 29 '18 at 6:56




















          • Can you clarify the Holder part?
            – Pratyush Sarkar
            Dec 29 '18 at 5:51






          • 1




            @Pratyush Sarkar I added something. See now.
            – Michael Rozenberg
            Dec 29 '18 at 6:46






          • 1




            Ah, thanks that helps a lot. Nice concise answer.
            – Pratyush Sarkar
            Dec 29 '18 at 6:56


















          Can you clarify the Holder part?
          – Pratyush Sarkar
          Dec 29 '18 at 5:51




          Can you clarify the Holder part?
          – Pratyush Sarkar
          Dec 29 '18 at 5:51




          1




          1




          @Pratyush Sarkar I added something. See now.
          – Michael Rozenberg
          Dec 29 '18 at 6:46




          @Pratyush Sarkar I added something. See now.
          – Michael Rozenberg
          Dec 29 '18 at 6:46




          1




          1




          Ah, thanks that helps a lot. Nice concise answer.
          – Pratyush Sarkar
          Dec 29 '18 at 6:56






          Ah, thanks that helps a lot. Nice concise answer.
          – Pratyush Sarkar
          Dec 29 '18 at 6:56













          1














          This does not seem the best, but using only Holder's inequality, we can get
          $$begin{eqnarray}
          S^{12}&le& (x+y+x+y)(y+y+x+x)(y+y+x+x)(z+y+z+y)\&&(y+z+z+y)(y+z+z+y)(z+w+z+w)(z+z+w+w)\&&(z+z+w+w)(x+w+x+w)(x+w+w+x)(x+w+w+x) = 2^{12}left(prod_{text{cyc}}(x+y)right)^3 =2^{12},
          end{eqnarray}$$          for $S= sqrt[3]{xyz}+sqrt[3]{yzw}+sqrt[3]{zwx}+sqrt[3]{wxy}.$






          share|cite|improve this answer





















          • Very nice solution! +1
            – Michael Rozenberg
            Dec 30 '18 at 9:53


















          1














          This does not seem the best, but using only Holder's inequality, we can get
          $$begin{eqnarray}
          S^{12}&le& (x+y+x+y)(y+y+x+x)(y+y+x+x)(z+y+z+y)\&&(y+z+z+y)(y+z+z+y)(z+w+z+w)(z+z+w+w)\&&(z+z+w+w)(x+w+x+w)(x+w+w+x)(x+w+w+x) = 2^{12}left(prod_{text{cyc}}(x+y)right)^3 =2^{12},
          end{eqnarray}$$          for $S= sqrt[3]{xyz}+sqrt[3]{yzw}+sqrt[3]{zwx}+sqrt[3]{wxy}.$






          share|cite|improve this answer





















          • Very nice solution! +1
            – Michael Rozenberg
            Dec 30 '18 at 9:53
















          1












          1








          1






          This does not seem the best, but using only Holder's inequality, we can get
          $$begin{eqnarray}
          S^{12}&le& (x+y+x+y)(y+y+x+x)(y+y+x+x)(z+y+z+y)\&&(y+z+z+y)(y+z+z+y)(z+w+z+w)(z+z+w+w)\&&(z+z+w+w)(x+w+x+w)(x+w+w+x)(x+w+w+x) = 2^{12}left(prod_{text{cyc}}(x+y)right)^3 =2^{12},
          end{eqnarray}$$          for $S= sqrt[3]{xyz}+sqrt[3]{yzw}+sqrt[3]{zwx}+sqrt[3]{wxy}.$






          share|cite|improve this answer












          This does not seem the best, but using only Holder's inequality, we can get
          $$begin{eqnarray}
          S^{12}&le& (x+y+x+y)(y+y+x+x)(y+y+x+x)(z+y+z+y)\&&(y+z+z+y)(y+z+z+y)(z+w+z+w)(z+z+w+w)\&&(z+z+w+w)(x+w+x+w)(x+w+w+x)(x+w+w+x) = 2^{12}left(prod_{text{cyc}}(x+y)right)^3 =2^{12},
          end{eqnarray}$$          for $S= sqrt[3]{xyz}+sqrt[3]{yzw}+sqrt[3]{zwx}+sqrt[3]{wxy}.$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 30 '18 at 8:59









          SongSong

          6,640319




          6,640319












          • Very nice solution! +1
            – Michael Rozenberg
            Dec 30 '18 at 9:53




















          • Very nice solution! +1
            – Michael Rozenberg
            Dec 30 '18 at 9:53


















          Very nice solution! +1
          – Michael Rozenberg
          Dec 30 '18 at 9:53






          Very nice solution! +1
          – Michael Rozenberg
          Dec 30 '18 at 9:53




















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