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!

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.

    x>0, y>0, z>0

    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:

    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:

    \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)


    \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:
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook