1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Prove inequality

  1. Jan 4, 2012 #1
    I found this problem the other day, seems interesting but I am still not sure about the solution
    Anybody can help

    x, y, z are numbers with

    x+y+z=1 and 0<x,y,z<1

    prove that

    sqrt(xy/(z+xy))+sqrt(yz/(x+yz))+sqrt(xz/(y+xz))<=3/2 ("<=" means less or equal)
     
    Last edited: Jan 4, 2012
  2. jcsd
  3. Jan 4, 2012 #2

    mathman

    User Avatar
    Science Advisor
    Gold Member

    I haven't worked it through. However, it looks like the max is assumed when x=y=z=1/3. In that case each term in the sum = 1/2.
     
  4. Jan 4, 2012 #3
    You are right I did that part already and think it's not enough
     
  5. Jan 4, 2012 #4

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    What do you mean, it is not enough?? Why do you think that??
     
  6. Jan 4, 2012 #5
    It should be proved for all other combinations of x,y and z.
    Proving only for x,y and z equal 1/3 is incomplete
     
  7. Jan 4, 2012 #6

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    But the expression reaches a maximum when x=y=z=1/3, so doesn't that tell you anything?
     
  8. Jan 5, 2012 #7



    Devil's Advocate:

    ** You claim that the maximum is reached when x = y = z = 1/3, but
    it has not been shown by anyone in this thread that it is (or referenced to
    another place as already known to be).

    So . . . . . I am not convinced that the expression could not be greater than 3/2.


    mathman stated that "it looks like the max..." <---- Not concrete

    When JennyPA stated that she "did that part already" in response
    to mathman, she may have just referred to plugging in 1/3 for each
    variable to see that particular sum is 3/2 , but had no knowledge of
    whether that selection makes a maximum or not.
     
    Last edited: Jan 5, 2012
  9. Jan 5, 2012 #8

    mathman

    User Avatar
    Science Advisor
    Gold Member

    I suggest that someone use some elementary calculus and see of the max is where it appears to be. To avoid partial derivatives, hold x constant and see if the max is at y=z. If that works, the rest should be easy enough.
     
  10. Jan 5, 2012 #9
    There was a previous discussion on this forum about this hypothesis:

    If f(x,y,z) is continuous and symmetric, that is, f(x,y,z) = f(y,z,x) = f(z,x,y), then if that function has a global maximum at f(x*,y*,z*), then x*=y*=z*

    That discussion didn't include a proof of that, though.
     
  11. Jan 5, 2012 #10

    pwsnafu

    User Avatar
    Science Advisor

    IIRC it was a function had to have a unique global maximum.
     
  12. Jan 5, 2012 #11
    Good point: if it's unique, then obviously x*=y*=z*
     
  13. Jan 6, 2012 #12
    Just a little algebra and the each term can eliminate a variable: xy/(z + xy) = xy/[(1-x)(1-y)]
    Likewise for the other 2 terms.
     
  14. Feb 25, 2012 #13
    I am still working on this problem and has no clue how to proceed from here.
     
  15. Feb 25, 2012 #14

    Office_Shredder

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    The best suggestion kind of got lost in the noise here. Take derivatives and use Lagrange multipliers/whatever calculus method you feel most comfortable with to find the maximum of the function
     
  16. Feb 25, 2012 #15
    The problem is I am not suppose to use calculus for this problem. This is the algebra course
     
  17. Feb 25, 2012 #16
    xy/[(1-x)(1-y)] when you eliminate z, then this expression is always 1 when x+y = 1
     
  18. Feb 26, 2012 #17
    how you eliminate z????
    you can replace z with 1-x-y but can't see what you did
     
  19. Feb 26, 2012 #18
    That's what I did, Then 1-x-y+xy =(1-x)(1-y)
     
  20. Feb 26, 2012 #19
    I understand that part I did that but can't get the statment x+y=1
     
  21. Feb 26, 2012 #20
    since z is no longer a part of this term, I looked at values for x and y to see where the max for the term is. The restrictions of x + y + z = 1 then x + y can't exceed 1.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Prove inequality
  1. Proving Inequalities (Replies: 1)

  2. An inequality (Replies: 11)

Loading...