
Elementary Math Fractions, Percentages, Word Problems, Equations, Inequations, Factorization, Expansion 
 LinkBack  Thread Tools  Display Modes 
October 18th, 2017, 09:19 AM  #1 
Senior Member Joined: Dec 2015 From: Earth Posts: 194 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: 84 Thanks: 14 Math Focus: Algebra, Cryptography 
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: 276 Thanks: 141 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: 194 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: 6,436 Thanks: 562 
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. 

Tags 
inequality 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Triangle Inequality: Prove Absolute Value Inequality  StillAlive  Calculus  5  September 3rd, 2016 12:45 AM 
An inequality  Dacu  Algebra  8  June 20th, 2014 09:20 PM 
inequality  dupek707  Number Theory  4  November 3rd, 2010 02:39 PM 
Inequality  Pavhard  Number Theory  4  October 18th, 2010 08:00 AM 
An inequality  bigli  Calculus  1  June 8th, 2010 07:44 AM 