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Show that $(n!)^2>n^n$ for $n \in \Bbb Z_+$. I tried induction but I was getting stuck in the last step. Please help. Hints only, please.

David
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Adienl
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  • This is only true for $n>2$. – user133281 May 08 '14 at 05:53
  • I know that's why I mentioned it in the title. – Adienl May 08 '14 at 05:56
  • Hint: write $(n!)^2$ as $(1\times n)(2\times(n-1))\cdots(n\times1)$. Can you show that the product in each bracket is at least $n$? I wouldn't try induction because I can't see any simple relation between $n^n$ and $(n+1)^{n+1}$. – David May 08 '14 at 06:00
  • @ David: what was your motivation behind that approach? – Adienl May 08 '14 at 06:13
  • By your method, I need to prove that $a(n-b)>=n$ with $a>=1, b>=0, and n> 2$. – Adienl May 08 '14 at 06:19
  • @david: I'm not able to prove it rigorously – Adienl May 08 '14 at 06:29
  • Re: $a(n-b)$ comment, you don't need to prove it for all $a,b$ because $a,b$ are closely related to each other. Have another look at my previous comment and if it's not clear, write out a few more terms. – David May 08 '14 at 06:40

4 Answers4

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We have $$(n!)^2 = \prod_{a=1}^n a(n+1-a) > \prod_{a=1}^n n = n^n,$$ because $a(n+1-a) \geq n$ for all $1 \leq a \leq n$. The latter inequality is equivalent to $(n-a)(a-1) \geq 0$ which is obviously true and also strict when $a=2$, $n>2$.

user133281
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HINT:

Write $1$ to $n$ in a line and below that write $n$ to $1$. Now multiply the two no.s in each columns.

  • But I can't prove it rigorously after that – Adienl May 08 '14 at 06:31
  • You have to just verify $(1+n-k)k \geq n$ for $k= 1,...,n$. Do the derivative wrt $k$ to check that maxima occurs at $(n+1)/2$ and the function monotonically decreases on both the sides of $k=(n+1)/2$ (as you go away from this $k$). Now you just need to verify that the extreme values (for $k= 1 , n$) are $\geq n$(which they are). –  May 08 '14 at 06:42
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Okay since you've said you tried induction, I'm assuming you reached the step

$(k!)^2 * (k+1)^2 > k^k*(k+1)^2 $

Now if you can show that

$k^k*(k+1)^2>(k+1)^{k+1}$

then you're done.

Why not try keeping the (k+1) terms in the right hand side?

Now we arrive at

$k^k>(k+1)^{k-1}$

Both the terms we have are in the form of products of integers, it might be easier if only one side is a product.(Why?)

So we can rewrite it again as

$1>\frac{1}{k}*(1+\frac{1}{k})^{k-1}$.

I hope now it is easy to see how to proceed.

Soham
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For a help on how to prove the problem by Induction suppose that the inequality is true for all $n>1$ , i.e, we have (for that $n$) $(n!)^2$ $>$ $n^n$ $\implies$ $\dfrac{(n!)^2}{n^n}$ $>$ $1$. Let $u_n$ $=$ $\dfrac{(n!)^2}{n^n}$. Then try to prove that $u_n$$_+$$_1$ $>$ $u_n$ for all $n>1$.

William Hilbert
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