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To prove that $(n!)^2>(n)^n$

That is to prove that $(n!)^2-(n)^n>0$

Now, $(n!)(n!)-(n)^n=[n.(n-1)(n-2)...][n.(n-1)(n-2)...]-[n.n.n...]$

$[n^2.(n-1)^2 . (n-2)^2 ...]-[n.n.n...]$

Comparing the first terms we get, $n^2>n$

Now we prove that all the other individual terms of $(n!)^2$ are greater than $n$

The general term is $(n-r)^2$

To prove that $(n-r)^2>n$, we subtract $n$ from $(n-r)^2 $

$(n-r)^2-n=n^2-2nr+r^2-n=n^2-(2r+1)n+r^2 $, but cannot proceed further, pls help, or this approach totally wrong?

Vikram
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1 Answers1

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Note that:

  • $n^n=\prod\limits_{i=1}^{n}n$
  • $(n!)^2=(1\times2\times\dots\times n)\times(1\times2\times\dots\times n)=\prod\limits_{i=1}^{n}i(n-i+1)$

So all you need to prove is:

  • $i=1 \implies i(n-i+1)=n$
  • $i=n \implies i(n-i+1)=n$
  • $1<i<n \implies i(n-i+1)>n$

The first two are obvious.

In order to prove the third one, simply show that for any arbitrary value of $n>2$, function $f(x)=x(n-x+1)-n=-x^2+(n+1)x-n$ is entirely positive in the range $x\in(1,n)$:

  • The roots of $f(x)$ are $x_{1,2}=\frac{-(n+1)\pm\sqrt{(n+1)^2-4n}}{-2}=\frac{n+1\pm\sqrt{n^2-2n+1}}{2}=\frac{n+1\pm(n-1)}{2}=1,n$
  • So $f(x)$ is either entirely positive or entirely negative in the range $x\in(1,n)$
  • It's sufficient to check a single value in that range, so pick $x=3$
  • $f(3)=3(n-2)-n$ is positive for any arbitrary $n>2$
  • So $f(x)$ is entirely positive in the range $x\in(1,n)$
barak manos
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  • is the method to prove the third is the only way to do it? – Vikram May 19 '14 at 14:04
  • @Vikram: Hard to say. There are probably easier ways to do it, as it seems pretty obvious that the product of two positive numbers is always larger than their sum minus one. – barak manos May 19 '14 at 14:06
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    When plugging in $i=2$ don't we get the condition that $n>2$? – DanZimm May 19 '14 at 14:38
  • One may also remember that $(x-a)(b-x)$ has maximum at $x=(a+b)/2$ (it's a parabola...) and minimum on $[a,b]$ is found at $a$ and $b$. – Jean-Claude Arbaut May 19 '14 at 14:39
  • @DanZimm: I do not understand your first comment, and I do not think that your second comment is correct. When $A=-1$ and $C=-n$, it should be $-4n$. – barak manos May 19 '14 at 14:46
  • @barakmanos oop ya my second comment is incorrect, nonetheless you need the condition, in order for this inequality to be true, that $n>2$ – DanZimm May 19 '14 at 14:50
  • @DanZimm: Yes, true, I just noticed that. But changing the condition of $n>1$ to $n>2$ is perfectly "aligned" with the proof, as we chose $x=2$, while proving correctness for $x\in(1,n)$. Thank you for the correction. – barak manos May 19 '14 at 14:52
  • The inequality $(n!)^2 > n^n$ simply isn't true for $n=2$. Following your outline we get that the terms are equal, not $>$. – DanZimm May 19 '14 at 14:56
  • @DanZimm: Gotcha, fixed. Thanks again. – barak manos May 19 '14 at 14:58