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The inequality is: $$n! > n^{\frac{n}{2}}$$ This is what I've tried:

$n=3 \Rightarrow 6>3^{\frac32}\\$

Assume the inequality holds for some $k\in\Bbb{N}: k! > k^{\frac{k}{2}}$

Prove the inequality holds for $(k+1)! > (k+1)^{\frac{k+1}{2}}$ Then:

$\begin{align*} (k+1)! &> (k+1)^{\frac{k+1}{2}} \\ (k+1) \cdot k! &> (k+1)^{\frac{k+1}{2}} \\ \end{align*}$

Now using the inductive hypothesis:

$\begin{align*} k! &> k^{\frac{k}{2}} \\ (k+1) \cdot k! &> (k+1) \cdot k^{\frac{k}{2}} \end{align*} $

We get:

$\begin{align*}(k+1) \cdot k^{\frac{k}{2}} &> (k+1)^{\frac{k+1}{2}} \end{align*} $

And this is where I got to. Any help is appreciated.

Dzemal Poparic
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    If you are trying to prove that something holds for $n\gt 2$ then the base case is $n=3$. In fact, $1!=\sqrt{1}$ so the inequality doesn't hold for $n=1$. – John Douma Nov 09 '24 at 23:10
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    $3$ is not greater than $3^\frac{3}{2}$ – jjagmath Nov 09 '24 at 23:52
  • Start with $n=3$ : $3^{3\over2} =3\sqrt{3} < 3\sqrt{4} = 3\times 2 =6! $ – Maksim Nov 09 '24 at 23:56
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    This question is similar to: https://math.stackexchange.com/questions/640602/show-that-if-n2-then-n2nn – If you believe it’s different, please [edit] the question, make it clear how it’s different and/or how the answers on that question are not helpful for your problem. – Gerry Myerson Nov 10 '24 at 00:14
  • See also https://math.stackexchange.com/questions/923990/proving-that-if-n2-then-nnn-2-using-induction and https://math.stackexchange.com/questions/339036/how-to-prove-this-inequality-with-factorials-nn-frac-n2 and probably many more. – Gerry Myerson Nov 10 '24 at 00:15

3 Answers3

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Try

if $n>2$ let $1\le k \le n-1$ then either $k$ or $n+1-k$ is $\ge n$. So $k(n+1-k) \ge n$

taking product we get $(n-1)!n! \ge n^{n-1}$ so $n!^2 \ge n^{n}$ thus $n! > n^{n/2}$

titanicrose
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  • How can k<=n-1 and >= n? and n+1-k>=n only if k<=1. It's true that k(n+1-k)>=n and >n if k>1 but your proof of that doesn't work. What are you taking the product of to get the (n-1)! and how did >= turn into > at the end? – Martin Rattigan Nov 10 '24 at 02:22
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The problem is simplified by taking logarithms. For $n!$ we have from Stirling's approximation:

$$ \ln n! > n \ln n - n $$

Therefore if we can show:

$$ n \ln n - n > \frac{n}{2} \ln n $$

Or:

$$ \ln n > 2 $$

then we're done. We can show it holds for n > 8 so for $n > 8$ we conclude $n! > n^{\frac{n}{2}}$. The remaining $2 < n < 8$ can be verified separately.

dkm
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Assume $n>2$ $$n!=\prod_1^ni$$ For even $n>2$, say $n=2m$ $$n!=\prod_1^mi\prod_1^m(n+1-i)=\prod_1^mi(n+1-i)$$ and for odd $n$, say $n=2m+1$ $$n!=(m+1)\prod_1^mi(n+1-i)$$ Let $f(i)=i(n+1-i)$, then $$df/di=n+1-2i$$ which is positive for $i\in [1,(n+1)/2)$ hence for $i\in [1,m]$.

When $i=1$, $f(i)=n$, so $f(i)>n$ for $i\in (1,m]$.
Hence $$\prod_1^mf(i)=\prod_1^mi(n+1-i)>n^m\text{ if }m>1$$

That is the required result for even $n$.

For odd $n$ we need to show $m+1>\sqrt n$

From a.m$\geqslant$g.m. $$(m+1)/2\geqslant\sqrt m$$ $$m+1\geqslant\sqrt{4m}>\sqrt{2m+1}\text{ since }m\geqslant 1$$

And that completes the required result for odd $n$.

Of course, there's not much mathematical induction apparent, but then you can always show $\phi(n)\Rightarrow\phi(n+1)$ if you can show $\phi(n+1)$.

Martin Rattigan
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