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Math Help - Inequalities

  1. #1
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    Inequalities

    Can anyone help me confirm if I have solved this correctly?

    Many thanks.

    Q.
    (1 + ax)n > 1 + anx, for a > 0, x > 0 & n \in \mathbb{N}

    Attempt: Step 1: For n = 1...
    (1 + ax)1 = 1 + ax & 1 + a(1)x = 1 + ax
    Since 1 + ax > 1 + ax, n is true.

    Step 2: Assume the statement is true for n = k, i.e. assume (1 + ax)k > 1 + akx.
    We must now show that the statement is true for n = k + 1,
    i.e. (1 + ax)k+1 > 1 + a(k + 1)x = 1 + akx + ax
    (1 + ax)k+1 = (1 + ax)(1 + ax)k > (1 + ax)(1 + akx)...((1 + ax)k > 1 + akx...assumed)
    If (1 + ax)(1 + akx) > 1 + akx + ax, then (1 + ax)k+1 > 1 + akx + ax
    (1 + ax)(1 + akx) > 1 + akx + ax
    if (1 + ax)(1 + akx) - (1 + akx + ax) > 0
    if akx2 + akx + ax + 1 - 1 - akx - ax > 0
    if akx2 > 0
    if x2[ak] > 0...true when a & x > 0 & n > 1
    (1 + ax)k+1 > 1 + akx + ax
    Therefore, the statement is true for n = k + 1, if true for n = k. Thus, the statement is true for all a, x & n > 1, n \in \mathbb{N}.
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  2. #2
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    Re: Inequalities

    Quote Originally Posted by GrigOrig99 View Post
    Can anyone help me confirm if I have solved this correctly?

    Many thanks.

    Q.
    (1 + ax)n > 1 + anx, for a > 0, x > 0 & n \in \mathbb{N}

    Attempt: Step 1: For n = 1...
    (1 + ax)1 = 1 + ax & 1 + a(1)x = 1 + ax
    Since 1 + ax > 1 + ax, n is true.

    Step 2: Assume the statement is true for n = k, i.e. assume (1 + ax)k > 1 + akx.
    We must now show that the statement is true for n = k + 1,
    i.e. (1 + ax)k+1 > 1 + a(k + 1)x = 1 + akx + ax
    (1 + ax)k+1 = (1 + ax)(1 + ax)k > (1 + ax)(1 + akx)...((1 + ax)k > 1 + akx...assumed)
    If (1 + ax)(1 + akx) > 1 + akx + ax,
    Why the "if" here?

    then (1 + ax)k+1 > 1 + akx + ax
    (1 + ax)(1 + akx) > 1 + akx + ax
    if (1 + ax)(1 + akx) - (1 + akx + ax) > 0
    if akx2 + akx + ax + 1 - 1 - akx - ax > 0
    if akx2 > 0
    if x2[ak] > 0...true when a & x > 0 & n > 1
    (1 + ax)k+1 > 1 + akx + ax
    Therefore, the statement is true for n = k + 1, if true for n = k. Thus, the statement is true for all a, x & n > 1, n \in \mathbb{N}.
    You have a whole series of "if"s here. Do you actually show any of those are true?

    The simplest way to do this is to actually mutiply (1+ ax)(1+ akx)). What do you get when you do that?
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  3. #3
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    Re: Inequalities

    Quote Originally Posted by GrigOrig99 View Post
    Can anyone help me confirm if I have solved this correctly?
    Q.
    (1 + ax)n > 1 + anx, for a > 0, x > 0 & n \in \mathbb{N}
    Attempt: Step 1: For n = 1...
    Since 1 + ax > 1 + ax, n is true.

    Step 2: Assume the statement is true for n = k, i.e. assume (1 + ax)k > 1 + akx.
    We must now show that the statement is true for n = k + 1,
    This is also known as Bernoull's inequality.

    You tend to over-do it and it is hard to read.
     \begin{align*}(1+ax)^{k+1}&=(1+ax)(1+ax)^k \\&\ge (1+ax)(1+akx) \\&=1+akx+ax+a^2kx\\ &\ge 1+akx+akx\\&=1+a(k+1)x\end{align*}
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  4. #4
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    Re: Inequalities

    I've mentioned this before, but I guess it bears repeating. The format I'm using to answer these questions is coming from the text book itself; an example of which I will provide in an attachment.

    Also, in the previous Inequalities question I posted, (1 +2x)(1 + 2kx) was solved by user 'Prove It' as follows:
    = 1 + 2kx + 2x + 2kx2

    Comparing his solution with that of (1 + ax)(1 + akx), shouldn't it be solved as:
    1 + akx + ax + akx2, rather than 1 + akx + ax + a2kx...?
    Attached Thumbnails Attached Thumbnails Inequalities-photo-2-.jpg  
    Last edited by GrigOrig99; September 12th 2012 at 08:13 AM.
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