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!

Help with the following inequality

  1. Aug 14, 2008 #1
    please help with the following inequality:

    .........................[(x+y)(y+z)(z+x) +xy+xz+yz]/xyz >=
  2. jcsd
  3. Aug 14, 2008 #2
    Re: inequalities

    That doesn't look like a very appealing inequality. Is there a condition for this inequality? I guess you could try dumbassing it by assuming x >= y >= z since the inequality is symmetric in x,y,z and then cross multiplying but I haven't figured out a nice solution yet.
  4. Aug 14, 2008 #3
    Re: inequalities

    thanks for the quick respond the only condition is that x>0,y>0,z>0
  5. Aug 14, 2008 #4


    User Avatar
    Science Advisor
    Homework Helper

    Re: inequalities

    What have you tried?
  6. Aug 15, 2008 #5
    Re: inequalities

    I would suggest, if x,y,z are positive, that you just get a common denominator and multiply the numerators out into expanded polynomials, then match up the terms.
  7. Aug 15, 2008 #6
    Re: inequalities

    Yeah just as I had guessed, the inequality is basically just multiplying out, moving everything to one side, and relying on the fact that the sum of positive real numbers is greater than 0.

    That is really a nasty inequality. It's symmetric but that doesn't make anything better. It's not homogeneous, so we can't come up with any additional constraints. It doesn't rely on any clever manipulations or other well-known inequalities (unless you count a simple albeit important fact). It's somewhat instructive but taken in perspective it's no where near as instructive as the other inequality you posed.
  8. Aug 15, 2008 #7
    Re: inequalities

    so what do we do now?
  9. Aug 15, 2008 #8
    Re: inequalities

    Give me a solid proof please
  10. Aug 16, 2008 #9
    Re: inequalities

    The change of variables:

    [tex]s = x + y + z, t = xyz, w = 1/x + 1/y + 1/z[/tex]

    might help. Note that [tex](x + y)(y + z)(x + z) = swt - t[/tex] and [tex]xy + xz + yz = tw[/tex]. The inequality becomes: [tex]sw + w - 1 \geq (s^2 + 2tw)/(stw - t)[/tex], if my calculations aren't off. Also note that [tex]sw \geq 1[/tex].
  11. Aug 16, 2008 #10
    Re: inequalities

    cellotim you mixed up worst now
  12. Aug 17, 2008 #11
    Re: inequalities

    Does this mean that I made a mistake or that I've confused you more?

    If you want to know how I would solve this problem for a class, I would sit down at Maple or Mathematica, cross-multiply the denominators, and have the computer symbolically expand each side (in x,y,z), then I would make sure that all the coefficients on the LHS are bigger than all the coefficients on the RHS. It would be a huge mess, but it would get it done. I don't know if you have access to any maths software, but that is how I would do it. (Of course you can do it all or partially by hand if you need to show that work. It wouldn't take more than 30-1hr.)
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook