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Math Help - mathematical induction

  1. #1
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    mathematical induction

    I have to prove this inequality using mathematical induction:

    (2)(4)(6).....(2n)
    ---------------- > 1 + (1/3) + (1/5) + .... + 1/(2n-1).
    (1)(3)(5)....(2n-1)

    I'm given that for positive reals a1, a2, a3,....,an where n >= 2 that
    the cartesian product of (1 + ai) from i = 1 to n is greater than 1 + a1 + a2 + .... + an.


    I know how to do the inductions with equal signs, but I can't seem to replace all the terms before (n + 1) on the left hand side with the terms on the right hand side (after assuming the statement is true for all n) since they're not equal.

    How do you do this one?

    Don't worry about the base case.
    Thanks for any help.
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  2. #2
    MHF Contributor red_dog's Avatar
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    Let n=2:
    (1+a_1)(1+a_2)=1+a_1+a_2+a_1a_2>1+a_1+a_2
    so the inequality is true.
    Suppose the inequality is true for n and prove for n+1.
    That means (1+a_1)(1+a_2)\ldots(1+a_n)(1+a_{n+1})>1+a_1+a_2+\  ldots+a_n+a_{n+1}.
    Using the statement for n we have
    (1+a_1)(1+a_2)\ldots(1+a_n)(1+a_{n+1})>(1+a_1+a_2+  \ldots a_n)(1+a_{n+1})=
    =1+a_1+a_2+\ldots+a_n+a_{n+1}+a_{n+1}(1+a_1+a_2+\l  dots a_n)>1+a_1+a_2+\ldots+a_n+a_{n+1}
    Using the transitivity yelds the statement for n+1.
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  3. #3
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    I appreciate the help, but how do you prove the first inequality?
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by PvtBillPilgrim View Post
    I appreciate the help, but how do you prove the first inequality?
    what first inequality? you mean for n = 2? we need not "prove" that one, since it is obvious (we have one more positive term on the left side than we do on the right). or was that not what you were talking about?
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  5. #5
    MHF Contributor red_dog's Avatar
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    The first inequality can be written as
    \displaystyle (1+1)\left(1+\frac{1}{3}\right)\left(1+\frac{1}{5}  \right)\ldots\left(1+\frac{1}{2n-1}\right)>1+\frac{1}{3}+\frac{1}{5}+\ldots+\frac{1  }{2n-1}
    Apply the general inequality taking a_1=1,a_2=\frac{1}{3},\ldots,a_n=\frac{1}{2n-1}
    Then we have
    \displaystyle (1+1)\left(1+\frac{1}{3}\right)\left(1+\frac{1}{5}  \right)\ldots\left(1+\frac{1}{2n-1}\right)>1+1+\frac{1}{3}+\frac{1}{5}+\ldots+\frac  {1}{2n-1}>
    \displaystyle>1+\frac{1}{3}+\frac{1}{5}+\ldots+\fr  ac{1}{2n-1}
    and apply transitivity.
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  6. #6
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    Makes sense.

    Thanks a lot.
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  7. #7
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    Just a quick question.

    Does this last proof use mathematical induction or is it just a proof? It seems it doesn't prove the n + 1 case.
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  8. #8
    MHF Contributor red_dog's Avatar
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    The last proof is a particular case of the general case which was proved by induction.
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