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Two Analysis questions

  1. Sep 14, 2009 #1
    1) By writing a = (a+b) + (-b) use the Triangle Inequality to obtain |a| - |b| [tex]\leq[/tex] |a+b|. Then interchange a and b to show that ||a| - |b|| [tex]\leq[/tex] |a+b|.

    The replace b by -b to obtain ||a| - |b|| [tex]\leq[/tex] |a - b|.

    Okay. I am a bit lost.

    I started out by plugging in what they give me for a in the first line into the Triangle Inequality, but that just reduces back to the Triangle Inequality.

    I'm just not sure where to start.

    2) Let n be a natural number and [tex]a_{1}, a_{2}, ... a_{n} [/tex]be positive numbers. Prove that [tex] (1+a_{1})(1+a_{2})+...+(1+a_{n}) \geq 1+a_{1}+a_{2}+...+a_{n}.[/tex]

    and that

    [tex](a_{1}+a_{2}+...+a_{n})(a_{1}^{-1}+a_{2}^{-1}+...+a_{n}^{-1}) \geq n^{2})[/tex]

    For the first part of this problem I started out by expanding [tex](1+a_{1})(1+a_{2})+...+(1+a_{n})[/tex] for n = 3 and noticed that it would cancel all the terms on the right side making it a bunch of terms greater than or equal to zero, I just couldn't generalize it for n and n+1.

    I haven't started on the second part yet.

  2. jcsd
  3. Sep 14, 2009 #2


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    |a+b+(-b)| = |(a-b) + b| <= |a-b| + |b|

    Can you see how to finish that off?

    Do it by induction. If the result holds for n terms, does it hold for n+1? (1+a1)*...*(a+an+1) = (1+a2)....(1+an)*1 + (1+a2)....(1+an)*an+1. Use the inductive hypothesis after doing this distribution
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