$x,y,z > 0$. If $xyz geq xy + yz + zx$ prove that $xyz geq 3(x+y+z)$

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Everything is in the title. But if possible, It would very nice if you gave indications rather than the solution.
Thanks in advance.







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  • From where does it come?
    – Dr. Sonnhard Graubner
    Aug 27 at 16:40










  • It's written " India 2001" but when I've checked their national Olympiad I couldn't find it ^^.
    – Frousse
    Aug 27 at 16:45










  • The "AM-GM inequality" may help.
    – John
    Aug 27 at 16:47










  • Hint: Use AM-HM inequality
    – Love Invariants
    Aug 27 at 16:48















up vote
1
down vote

favorite
1












Everything is in the title. But if possible, It would very nice if you gave indications rather than the solution.
Thanks in advance.







share|cite|improve this question






















  • From where does it come?
    – Dr. Sonnhard Graubner
    Aug 27 at 16:40










  • It's written " India 2001" but when I've checked their national Olympiad I couldn't find it ^^.
    – Frousse
    Aug 27 at 16:45










  • The "AM-GM inequality" may help.
    – John
    Aug 27 at 16:47










  • Hint: Use AM-HM inequality
    – Love Invariants
    Aug 27 at 16:48













up vote
1
down vote

favorite
1









up vote
1
down vote

favorite
1






1





Everything is in the title. But if possible, It would very nice if you gave indications rather than the solution.
Thanks in advance.







share|cite|improve this question














Everything is in the title. But if possible, It would very nice if you gave indications rather than the solution.
Thanks in advance.









share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 27 at 17:05









Michael Rozenberg

89.1k1581180




89.1k1581180










asked Aug 27 at 16:39









Frousse

164




164











  • From where does it come?
    – Dr. Sonnhard Graubner
    Aug 27 at 16:40










  • It's written " India 2001" but when I've checked their national Olympiad I couldn't find it ^^.
    – Frousse
    Aug 27 at 16:45










  • The "AM-GM inequality" may help.
    – John
    Aug 27 at 16:47










  • Hint: Use AM-HM inequality
    – Love Invariants
    Aug 27 at 16:48

















  • From where does it come?
    – Dr. Sonnhard Graubner
    Aug 27 at 16:40










  • It's written " India 2001" but when I've checked their national Olympiad I couldn't find it ^^.
    – Frousse
    Aug 27 at 16:45










  • The "AM-GM inequality" may help.
    – John
    Aug 27 at 16:47










  • Hint: Use AM-HM inequality
    – Love Invariants
    Aug 27 at 16:48
















From where does it come?
– Dr. Sonnhard Graubner
Aug 27 at 16:40




From where does it come?
– Dr. Sonnhard Graubner
Aug 27 at 16:40












It's written " India 2001" but when I've checked their national Olympiad I couldn't find it ^^.
– Frousse
Aug 27 at 16:45




It's written " India 2001" but when I've checked their national Olympiad I couldn't find it ^^.
– Frousse
Aug 27 at 16:45












The "AM-GM inequality" may help.
– John
Aug 27 at 16:47




The "AM-GM inequality" may help.
– John
Aug 27 at 16:47












Hint: Use AM-HM inequality
– Love Invariants
Aug 27 at 16:48





Hint: Use AM-HM inequality
– Love Invariants
Aug 27 at 16:48











3 Answers
3






active

oldest

votes

















up vote
3
down vote



accepted










$$dfrac1x+dfrac1y+dfrac1z le 1$$



$$3left(dfrac1xy+dfrac1yz+dfrac1zxright)leleft(dfrac1x+dfrac1y+dfrac1zright)^2le 1$$






share|cite|improve this answer
















  • 1




    Well, thank you but that's pretty much the whole solution, don't you think ? ^^
    – Frousse
    Aug 27 at 16:59

















up vote
3
down vote













Let $1/x=a, 1/y=b, 1/z=c$.



Then the problem is: given $a,b,c>0$ such that $a+b+cleq 1$, show that $ab+bc+caleq 1/3$.



This is simple:
$1geq (a+b+c)^2geq 3(ab+bc+ca)$.






share|cite|improve this answer



























    up vote
    1
    down vote













    Since $$1geqfracxy+xz+yzxyz,$$ it's enough to prove that
    $$xyzleft(fracxy+xz+yzxyzright)^2geq3(x+y+z)$$ or
    $$(xy+xz+yz)^2geq3xyz(x+y+z)$$ or
    $$sum_cycz^2(x-y)^2geq0.$$






    share|cite|improve this answer




















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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes








      up vote
      3
      down vote



      accepted










      $$dfrac1x+dfrac1y+dfrac1z le 1$$



      $$3left(dfrac1xy+dfrac1yz+dfrac1zxright)leleft(dfrac1x+dfrac1y+dfrac1zright)^2le 1$$






      share|cite|improve this answer
















      • 1




        Well, thank you but that's pretty much the whole solution, don't you think ? ^^
        – Frousse
        Aug 27 at 16:59














      up vote
      3
      down vote



      accepted










      $$dfrac1x+dfrac1y+dfrac1z le 1$$



      $$3left(dfrac1xy+dfrac1yz+dfrac1zxright)leleft(dfrac1x+dfrac1y+dfrac1zright)^2le 1$$






      share|cite|improve this answer
















      • 1




        Well, thank you but that's pretty much the whole solution, don't you think ? ^^
        – Frousse
        Aug 27 at 16:59












      up vote
      3
      down vote



      accepted







      up vote
      3
      down vote



      accepted






      $$dfrac1x+dfrac1y+dfrac1z le 1$$



      $$3left(dfrac1xy+dfrac1yz+dfrac1zxright)leleft(dfrac1x+dfrac1y+dfrac1zright)^2le 1$$






      share|cite|improve this answer












      $$dfrac1x+dfrac1y+dfrac1z le 1$$



      $$3left(dfrac1xy+dfrac1yz+dfrac1zxright)leleft(dfrac1x+dfrac1y+dfrac1zright)^2le 1$$







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Aug 27 at 16:48









      Sergic Primazon

      40215




      40215







      • 1




        Well, thank you but that's pretty much the whole solution, don't you think ? ^^
        – Frousse
        Aug 27 at 16:59












      • 1




        Well, thank you but that's pretty much the whole solution, don't you think ? ^^
        – Frousse
        Aug 27 at 16:59







      1




      1




      Well, thank you but that's pretty much the whole solution, don't you think ? ^^
      – Frousse
      Aug 27 at 16:59




      Well, thank you but that's pretty much the whole solution, don't you think ? ^^
      – Frousse
      Aug 27 at 16:59










      up vote
      3
      down vote













      Let $1/x=a, 1/y=b, 1/z=c$.



      Then the problem is: given $a,b,c>0$ such that $a+b+cleq 1$, show that $ab+bc+caleq 1/3$.



      This is simple:
      $1geq (a+b+c)^2geq 3(ab+bc+ca)$.






      share|cite|improve this answer
























        up vote
        3
        down vote













        Let $1/x=a, 1/y=b, 1/z=c$.



        Then the problem is: given $a,b,c>0$ such that $a+b+cleq 1$, show that $ab+bc+caleq 1/3$.



        This is simple:
        $1geq (a+b+c)^2geq 3(ab+bc+ca)$.






        share|cite|improve this answer






















          up vote
          3
          down vote










          up vote
          3
          down vote









          Let $1/x=a, 1/y=b, 1/z=c$.



          Then the problem is: given $a,b,c>0$ such that $a+b+cleq 1$, show that $ab+bc+caleq 1/3$.



          This is simple:
          $1geq (a+b+c)^2geq 3(ab+bc+ca)$.






          share|cite|improve this answer












          Let $1/x=a, 1/y=b, 1/z=c$.



          Then the problem is: given $a,b,c>0$ such that $a+b+cleq 1$, show that $ab+bc+caleq 1/3$.



          This is simple:
          $1geq (a+b+c)^2geq 3(ab+bc+ca)$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Aug 27 at 16:50









          A. Pongrácz

          4,415725




          4,415725




















              up vote
              1
              down vote













              Since $$1geqfracxy+xz+yzxyz,$$ it's enough to prove that
              $$xyzleft(fracxy+xz+yzxyzright)^2geq3(x+y+z)$$ or
              $$(xy+xz+yz)^2geq3xyz(x+y+z)$$ or
              $$sum_cycz^2(x-y)^2geq0.$$






              share|cite|improve this answer
























                up vote
                1
                down vote













                Since $$1geqfracxy+xz+yzxyz,$$ it's enough to prove that
                $$xyzleft(fracxy+xz+yzxyzright)^2geq3(x+y+z)$$ or
                $$(xy+xz+yz)^2geq3xyz(x+y+z)$$ or
                $$sum_cycz^2(x-y)^2geq0.$$






                share|cite|improve this answer






















                  up vote
                  1
                  down vote










                  up vote
                  1
                  down vote









                  Since $$1geqfracxy+xz+yzxyz,$$ it's enough to prove that
                  $$xyzleft(fracxy+xz+yzxyzright)^2geq3(x+y+z)$$ or
                  $$(xy+xz+yz)^2geq3xyz(x+y+z)$$ or
                  $$sum_cycz^2(x-y)^2geq0.$$






                  share|cite|improve this answer












                  Since $$1geqfracxy+xz+yzxyz,$$ it's enough to prove that
                  $$xyzleft(fracxy+xz+yzxyzright)^2geq3(x+y+z)$$ or
                  $$(xy+xz+yz)^2geq3xyz(x+y+z)$$ or
                  $$sum_cycz^2(x-y)^2geq0.$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Aug 27 at 17:01









                  Michael Rozenberg

                  89.1k1581180




                  89.1k1581180



























                       

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