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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


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    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


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    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


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    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


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    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


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    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


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    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.
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