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October 18th, 2017, 09:19 AM  #1 
Senior Member Joined: Dec 2015 From: Earth Posts: 181 Thanks: 23  n! inequality
How to prove if $\displaystyle \; {(\frac{n}{3})}^{n}\leq n! $

October 18th, 2017, 09:32 AM  #2 
Member Joined: Aug 2011 From: Nouakchott, Mauritania Posts: 69 Thanks: 6 
Salam ! Did you try to show this inequality by induction ? You may use the Binomial Expansion. 
October 18th, 2017, 09:56 AM  #3 
Senior Member Joined: Sep 2016 From: USA Posts: 223 Thanks: 120 Math Focus: Dynamical systems, analytic function theory, numerics 
It isn't a true statement so it would be hard to prove.

October 18th, 2017, 10:29 AM  #4 
Senior Member Joined: Dec 2015 From: Earth Posts: 181 Thanks: 23  In my book it says true, but I want to see how to get there? And I forgot to write $\displaystyle n \in N$. Last edited by skipjack; October 18th, 2017 at 12:14 PM. 
October 18th, 2017, 03:00 PM  #5 
Global Moderator Joined: May 2007 Posts: 6,397 Thanks: 546 
An approximate proof would start using Stirling's formula: $\displaystyle n!\simeq \sqrt {2\pi n}(\frac{n}{e})^n>(\frac{n}{3})^n$ Checking on the derivation of Stirling's formula. $\displaystyle \sqrt {2\pi n}(\frac{n}{e})^n$ is a lower bound for n!. Last edited by mathman; October 19th, 2017 at 02:02 PM. 

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