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Prove (a+b)!<a!b!2^(a+b) help please!

  1. Sep 12, 2011 #1
    The actual question is prove that [tex] |\alpha|!\le n^{|\alpha|}\alpha![/tex] where
    [tex]\alpha=(\alpha_1,.....\alpha_n)[/tex] is a multi-index (all non-negative) and [tex]
    |\alpha|=\alpha_1+\cdots +\alpha_n [/tex] and [tex]\alpha!=\alpha_1!\cdots \alpha_n! [/tex] so I am trying to do it by induction on the number of elements [tex]n[/tex] in [tex]\alpha[/tex]...so I am trying to prove that [tex](a+b)!<2^{a+b}a!b! [/tex] I have tried to do this by induction on the value of b (the inequality is obvious for b=0 or 1), and other ways, but nothing is working (been trying for close to a week).

    Can someone please help? :)

    (ps. how do I make it so that after I write in latex it doesn't skip a line like that?)
     
  2. jcsd
  3. Sep 12, 2011 #2

    micromass

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    Are you allowed to use Stirling approximation??

    By the way, use [itex ] if you don't want newlines.
     
  4. Sep 12, 2011 #3

    micromass

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    Anyway, if you're not allowed to use Stirling approximation, just notice that your inequality is equivalent to

    [tex]\binom{a+b}{a}<2^{a+b}[/tex]

    Now you can use a combinatorial argument.
     
  5. Sep 12, 2011 #4
    I believe I am allowed to use stirling's approximation, can you suggest a way? (it's only approximate for large n).

    Anyway, I will try the other way in the mean time, thanks.
     
  6. Sep 12, 2011 #5
    Never mind, I got it! You were a huge help, thank you!
     
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