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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

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    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

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    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

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    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

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    That's right.

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

    dextercioby

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    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
     
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