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Proving a desired inequality

  1. Jul 6, 2008 #1
    I'm interested in proving the following inequality, given the following constraints. This is for personal gain, and not for homework. I'm not even sure if I can prove this without additional constraints. If you can solve it by adding constraints, please let me know what options I have for adding constraints.

    Given:
    0<a<1
    x>0, y>0, z>0
    x>y

    Prove:[tex]\frac{x-y}{(1+x)(1+y)} < \frac{z*a}{1+y+z}[/tex]

    Thanks so much for your help!
     
    Last edited: Jul 6, 2008
  2. jcsd
  3. Jul 6, 2008 #2
    I believe it's proven wrong with the example, (a,x,y,z)=(1/2,2,1,1).
     
  4. Jul 6, 2008 #3
    You are indeed correct.

    How about we add the constraint:
    [tex]y+z>x[/tex]

    What about now?
     
  5. Jul 7, 2008 #4
    Well y+z>x really just means z>x in your case, but I'll try to prove it:

    [tex]
    \frac {x-y}{(1+x)(1+y)}<\frac {z*a}{1+y+z}[/tex]

    Fallowing constraints, we set (a,x,y,z) to (1/2,2,1,2)

    [tex]

    \frac {2-1}{(1+2)(1+1)}<\frac {2*1/2}{1+1+2}[/tex]

    [tex]\frac {1}{6}<\frac {1}{4}[/tex]

    This proves true.
    Assume [tex](a,x,y,z)\longrightarrow(a+1,x+1,y+1,z+1)[/tex]; plug in the original set.

    [tex]\frac {(x+1)-(y+1)}{(1+(x+1))(1+(y+1))}<\frac {(z+1)(a+1)}{1+(y+1)+(z+1)}[/tex]

    [tex]\frac {x-y}{(x+2)(y+2)}<\frac {(z+1)(a+1)}{y+z+3}[/tex]

    Now plug in, and simplify:

    [tex]\frac {2-1}{(2+2)(1+2)}<\frac {(2+1)(1/2+1)}{1+2+3}[/tex]

    [tex]\frac {1}{12}<\frac {9}{12}[/tex]

    I don't know if this is formal, but I'd call it adequate.
     
  6. Jul 7, 2008 #5
    I think it works well for larger values of x,y,z. But when x,y,z are small, for example, between 0 and 1, I think your method won't work.
     
  7. Jul 7, 2008 #6
    I should have set (a,x,y,z) to (d,2d,d,3d) and (1-d,2d,d,3d).
    Anyways, you have to fix your constraints...
     
  8. Jul 7, 2008 #7
    I can't exactly "fix my constraints". I'm trying to prove something as generally as possible. Can you think of any constraints I can add that would easily establish the desired result? Because I can't.
     
  9. Jul 7, 2008 #8
    Here are some constraints that I think work:
    x>y
    0<y<1
    a>6
    z>2/35
     
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