Results 1 to 2 of 2

Math Help - Equality

  1. #1
    Newbie
    Joined
    Feb 2009
    Posts
    5

    Equality

    What are the min and max values of:

    \frac{1}{1+x+xy}+\frac{1}{1+y+yz}+\frac{1}{1+z+zx}

    for reals x,y,z such that xyz=1?

    EDIT: I now see that 1 is the only possible value if only xyz=1, but I don't know how to reduce it to show that fact. Anyone?
    Last edited by Jone; February 20th 2009 at 08:07 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor red_dog's Avatar
    Joined
    Jun 2007
    From
    Medgidia, Romania
    Posts
    1,252
    Thanks
    5
    \frac{1}{1+x+xy}+\frac{1}{1+y+yz}+\frac{1}{1+z+zx}  =

    =\frac{1}{1+x+\frac{1}{z}}+\frac{1}{y+yz}+\frac{1}  {1+z+zx}=

    =\frac{z}{1+z+zx}+\frac{1}{1+y+yz}+\frac{1}{1+z+zx  }=

    =\frac{1+z}{1+z+zx}+\frac{1}{1+y+yz}=

    =\frac{1+z}{1+z+\frac{1}{y}}+\frac{1}{y+yz}=

    =\frac{y+yz}{1+y+yz}+\frac{1}{1+y+yz}=

    =\frac{1+y+yz}{1+y+yz}=1

    So, the expression is constant.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 4
    Last Post: December 17th 2011, 11:48 AM
  2. Equality
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: May 13th 2010, 02:30 PM
  3. Equality !!
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: January 31st 2010, 06:33 AM
  4. equality
    Posted in the Algebra Forum
    Replies: 2
    Last Post: September 18th 2008, 12:43 PM
  5. Set equality
    Posted in the Discrete Math Forum
    Replies: 5
    Last Post: January 24th 2008, 04:04 PM

Search Tags


/mathhelpforum @mathhelpforum