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November 29th, 2017, 09:03 AM   #1
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Another factorial inequality

Prove that $\displaystyle n!^2 \geq n^n$ for $\displaystyle n\in N$
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November 29th, 2017, 02:45 PM   #2
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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.
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December 28th, 2017, 09:39 AM   #3
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Originally Posted by LearnerLearner View Post
Use induction and then everything can be solved
This may be the stupidest answer ever on this or any other math web site.
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December 28th, 2017, 09:49 AM   #4
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Originally Posted by JeffM1 View Post
This may be the stupidest answer ever on this or any other math web site.
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December 28th, 2017, 03:52 PM   #5
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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 03:54 PM.
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