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Even More Induction

  1. Jun 23, 2008 #1
    1. The problem statement, all variables and given/known data

    Show that

    [tex]\forall n \in \mathbb{N}: 3^{n} \geq n^{3}[/tex]

    3. The attempt at a solution

    (1) Show that it is true for n = 1:

    [tex]3^{1} \geq 1^{3}[/tex]

    (2) Show that if it is true for n = p, then it is true for n = p + 1:

    Assume that [tex]3^{p} \geq p^{3}[/tex]

    Now,

    [tex]3^{p+1} = 3 \cdot 3^{p} = 3^{p} + 3^{p} + 3^{p}[/tex]

    [tex](p+1)^{3} = p^{3} + 3p^{2} + 3p + 1[/tex]

    Given our assumption, we know that if it could be demonstrated that

    [tex]3^{p} + 3^{p} \geq 3p^{2} + 3p + 1[/tex]

    then we are done. From here, I'm not sure how to proceed. Should I pull some moves from analysis and argue that certain functions grow faster than others above a certain n? The last inequality is also a stronger criteria, but does not apply to p = 1 or p = 2, since 3^{p} was larger than p^{3}.
     
  2. jcsd
  3. Jun 23, 2008 #2

    dynamicsolo

    User Avatar
    Homework Helper

    Why not just compare

    [tex]3^{p} + 3^{p} + 3^{p}
    [/tex]

    term-by-term with

    [tex]
    p^{3} + 3p^{2} + (3p + 1)
    [/tex] ?

    You've assumed

    [tex]
    3^{p} \geq p^{3}
    [/tex]

    and, beyond some low value of p,

    [tex]
    3^{p} \geq 3p^{2}
    [/tex]

    and

    [tex]
    3^{p} \geq 3p+1
    [/tex].

    EDIT: the last two inequalities only fail for p = 1, but you've already shown that the proposed inequality works there. As for showing that the inequalities work for p>= 2,
    how about taking log base 3 of both sides?
     
    Last edited: Jun 23, 2008
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