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Inequality proof: sqrt(1+xi^2)-xi < 1, for xi > 0

  1. Jul 27, 2010 #1
    1. The problem statement, all variables and given/known data

    Show

    [tex]\sqrt{1+\xi^2}-\xi<1[/tex]

    for [tex]\xi>0[/tex]



    2. Relevant equations



    3. The attempt at a solution

    Is this correct way?

    [tex]\sqrt{1+\xi^2}-\xi<1[/tex]

    suppose

    [tex]\sqrt{1+\xi^2}-\xi\geq 1[/tex]

    [tex]\sqrt{1+\xi^2}\geq 1+\xi[/tex]

    [tex]1+\xi^2 \geq 1+2\xi+\xi^2[/tex]

    [tex]0 \geq \xi[/tex]

    contradiction

    so

    [tex]\sqrt{1+\xi^2}-\xi<1[/tex]

    for [tex]\xi>0[/tex]
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jul 27, 2010 #2

    vela

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    Re: Problem

    Yup. That works.

    Or you could just reverse your steps.

    Let ξ>0. Then 1+ξ2<1+ξ2+2ξ . . .
     
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