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Show that $$n! \leqslant (\frac{n+1}{2})^n \quad \hbox{for all } n \in \mathbb{N}$$

I know that it can be done by induction but I always find line where I do not know what to do next.

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Use the AM-GM inequality on the numbers $1,...,n$.

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    $\begingroup$ Very nice hint. +1 $\endgroup$ – Timbuc Oct 26 '14 at 17:29
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Hint: $$ (n!)^2 = 1\times n \times 2\times (n-1) \times \dots = \prod_{k=1}^n k(n+1-k) $$ then use $$ k\times (n+1 -k) = \left(\frac {n+1}2 + \frac {n+1}2 - k\right)\left(\frac {n+1}2 - \frac {n+1}2 + k\right) \\= \left(\frac {n+1}2\right)^2 - \left( \frac {n+1}2 - k\right)^2 \le \left(\frac {n+1}2\right)^2 $$ to get $$ (n!)^2 \le \left(\frac {n+1}2\right)^{2n} $$

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HINT:

$$\frac{r+n+1-r}2\ge\sqrt{r(n-r)}$$ for $1\le r\le n$

Set $r=1,2,\cdots,n-1,n$ and then multiply

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$$(n!)^2 = \prod_{k=1}^n k(n+1-k) \leq \prod_{k=1}^n \left(\dfrac{k+n+1-k}{2}\right)^2 = \left(\dfrac{n+1}{2}\right)^{2n}$$

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  • 2
    $\begingroup$ Words can be useful. $\endgroup$ – Ali Caglayan Oct 26 '14 at 17:40
  • $\begingroup$ @Alizter not always $\endgroup$ – Petite Etincelle Oct 26 '14 at 17:53

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