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 November 29th, 2017, 08:03 AM #1 Senior Member   Joined: Dec 2015 From: Earth Posts: 238 Thanks: 27 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: 6,556 Thanks: 600 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. Thanks from greg1313, Country Boy and JeffM1
December 28th, 2017, 08:39 AM   #3
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 Originally Posted by LearnerLearner Use induction and then everything can be solved
This may be the stupidest answer ever on this or any other math web site.

December 28th, 2017, 08:49 AM   #4
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 Originally Posted by JeffM1 This may be the stupidest answer ever on this or any other math web site.

 December 28th, 2017, 02:52 PM #5 Senior Member   Joined: Sep 2016 From: USA Posts: 414 Thanks: 228 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: $a_k = (n-k)(k+1) \quad k =0,1,\dots,n-1.$ 2. Now $n!^2 = a_0 a_1 \dots a_{n-1}$ 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 $a_k - n = (n-k)(k+1)-n = nk - k^2 -k.$ 4. Since $n \geq (k+1)k$, we have $(n-k-1) \geq 0$ 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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