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November 29th, 2017, 08:03 AM  #1 
Senior Member Joined: Dec 2015 From: Earth Posts: 224 Thanks: 26  Another factorial inequality
Prove that $\displaystyle n!^2 \geq n^n$ for $\displaystyle n\in N$

November 29th, 2017, 01:45 PM  #2 
Global Moderator Joined: May 2007 Posts: 6,511 Thanks: 584 
Mathematical induction can work. For n=1 or 2, equality. For n=3, 36 > 27. In general. $(n+1)!^2=(n+1)^2(n!)^2$ $(n+1)^{n+1}=(n+1)n^n(1+\frac{1}{n})^n<(n+1)n^ne<( n+1)^2(n!)^2$, for n > 2. 
December 28th, 2017, 08:39 AM  #3 
Senior Member Joined: May 2016 From: USA Posts: 1,030 Thanks: 420  
December 28th, 2017, 08:49 AM  #4 
Math Team Joined: Jul 2011 From: Texas Posts: 2,751 Thanks: 1401  
December 28th, 2017, 02:52 PM  #5 
Senior Member Joined: Sep 2016 From: USA Posts: 377 Thanks: 205 Math Focus: Dynamical systems, analytic function theory, numerics 
I would agree that this is a great exercise for using induction. In the interests of completeness, I'll add a sketch for a direct proof which doesn't require induction. 1. Note that $n!^2$ is a product of $2n$many factors which we can group into only $n$many factors by taking pairwise products by defining: \[a_k = (nk)(k+1) \quad k =0,1,\dots,n1. \] 2. Now $n!^2 = a_0 a_1 \dots a_{n1}$ and $n^n$ are each products of $n$many factors. It suffices to prove that $a_k \geq n$ for every $k$. 3. Fix $k$ and expand \[a_k  n = (nk)(k+1)n = nk  k^2 k. \] 4. Since $n \geq (k+1)k$, we have $ (nk1) \geq 0$ and since $k \geq 0$, this implies $a_k  n \geq 0$. Last edited by SDK; December 28th, 2017 at 02:54 PM. 

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