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!

Inequality math problem

  1. May 15, 2005 #1
    hello all:

    I have tried to solved the following problem, however, I was stucked. thanks for the help:

    x, y z are positive number,
    prove (x+y+z)(1/x+1/y+1/z) >=9,

    if so, how about (x+y+z+w)(1/x+1/y+1/z+1/w) >= 9.

    The following is what I've got:

    (x+y+z)(1/x+1/y+1/z) = (x+y+x)[(yz+xz+xy)/xyz]
    =[3(xyz) + (x^2*z+x^2*y+y^2*z+x*y^2+y*z^2+x*z^2)]/xyz
    = 3 + [(x^2*z+x^2*y+y^2*z+x*y^2+y*z^2+x*z^2)]/xyz

    Please help!!!!
     
  2. jcsd
  3. May 15, 2005 #2

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    It's no big deal

    [tex] \left(x+y+z\right)\left(x^{-1}+y^{-1}+z^{-1}\right)=3+\left[\left(xy^{-1}\right)+\left(yx^{-1}\right)\right]+\left[\left(xz^{-1}\right)+\left(zx^{-1}\right)\right]+\left[\left(yz^{-1}\right)+\left(zy^{-1}\right)\right][/tex]

    [tex] \geq 3+2+2+2=9 [/tex]

    Q.e.d.

    I hope u see why.

    Daniel.
     
  4. May 15, 2005 #3

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    Along the same lines,u can prove quite easily that

    [tex] \left(x+y+z+w\right)\left(x^{-1}+y^{-1}+z^{-1}+w^{-1}\right)\geq 16 [/tex]

    Daniel.
     
  5. May 15, 2005 #4
    Thanks, Daniel;

    How did you get the following line?
    >= 3 + 2 + 2 +2

    Many thanks
     
  6. May 15, 2005 #5

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    Well,take for example the first

    [tex] xy^{-1}+yx^{-1} [/tex] (1)

    Make the substitution [tex] xy^{-1}=a [/tex] (2).Then [tex] yx^{-1}=a^{-1} [/tex] (3)

    Therefore,(1) becomes [tex] a+a^{-1} [/tex] (4)

    And i say that [tex] a+a^{-1}\geq 2 [/tex] (5),for a>0.

    Can u prove it...?

    Daniel.
     
  7. May 15, 2005 #6
    Thanks, Daniel;

    I got it.
     
  8. May 15, 2005 #7
    Cool;

    a + 1/a = (a^2 + 1)/a >= 2 since
    (a-1)^2 >= 0
     
  9. May 15, 2005 #8

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    That's right.

    Daniel.
     
  10. May 16, 2005 #9
    Daniel, I prove it with AM-GM Inequalities.
     
  11. May 16, 2005 #10

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    I think you mean harmonic average.The geometric one is useless...

    Daniel.
     
  12. May 21, 2005 #11
    arithmatic mean >= harmonic mean
    i.e. A>=H
    i.e. A/H >= 1

    now
    . x+y+z+w = 4*A and
    1/x + 1/y + 1/z + 1/w = 4/H
    hence the product is (4*A)*4/H
    =16 * (A/H) >= 16*(1).
    QED
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Inequality math problem
  1. Inequality problem (Replies: 5)

  2. Inequation problem (Replies: 2)

  3. Inequality Problem (Replies: 30)

  4. Inequality problem (Replies: 2)

Loading...