Can Multinomial Coefficients Prove Factorial Inequalities?

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Homework Help Overview

The discussion revolves around proving the inequality |\alpha|! ≤ n^{|\alpha|}α! for a multi-index α, where |α| is the sum of its components and α! is the product of the factorials of its components. The original poster is attempting to prove related inequalities using induction.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster discusses using induction on the number of elements in α and also on the value of b in a related inequality. Some participants question the use of Stirling's approximation and suggest alternative combinatorial arguments.

Discussion Status

Participants are exploring different approaches to the problem, including induction and combinatorial methods. There is a mix of suggestions regarding the use of approximations and alternative reasoning, but no consensus has been reached on a definitive method.

Contextual Notes

There is mention of constraints regarding the use of Stirling's approximation and the original poster's uncertainty about formatting in LaTeX.

lackrange
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The actual question is prove that |\alpha|!\le n^{|\alpha|}\alpha! where
\alpha=(\alpha_1,...\alpha_n) is a multi-index (all non-negative) and <br /> |\alpha|=\alpha_1+\cdots +\alpha_n and \alpha!=\alpha_1!\cdots \alpha_n! so I am trying to do it by induction on the number of elements n in \alpha...so I am trying to prove that (a+b)!&lt;2^{a+b}a!b! 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?)
 
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Are you allowed to use Stirling approximation??

By the way, use [itex ] if you don't want newlines.
 
Anyway, if you're not allowed to use Stirling approximation, just notice that your inequality is equivalent to

\binom{a+b}{a}&lt;2^{a+b}

Now you can use a combinatorial argument.
 
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.
 
Never mind, I got it! You were a huge help, thank you!
 

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