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This question already has an answer here:

Prove that for $n\in\mathbb{N}$ $$n!\leq(\frac{n+1}{2})^n.$$ I've solved base case for $n=1$ $$1\leq(\frac{1+1}{2})^1=1$$ The second step I've made was that I assumed that $n!\leq(\frac{n+1}{2})^n$ And then I have $(\frac{n+2}{2})^{n+1}$. What do I need to do now?

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marked as duplicate by Petite Etincelle, mookid, Mark Bennet, Micah, Kirill Nov 3 '14 at 0:19

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Applying the AM-GM inequality you obtain $$\sqrt[n]n!=\sqrt[n]{1\cdot2\cdot\ldots\cdot(n-1)\cdot n}\overset{\text{AM-GM}}\le\frac{1+2+\ldots+(n-1)+n}{n}=\frac{n(n+1)}{2n}=\frac{n+1}{2}$$ Now raise both sides to the $n$-power to conclude.

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Just write that $$ (n!)^2 = \prod_{k=1}^n k(n+1-k) \le \prod_{k=1}^n \left[\frac 12 (k+(n+1-k))\right]^2 = \left(\frac{n+1}2\right)^{2n} $$

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The induction proof can be made to work; use the fact that

$$\begin{align*} \frac{\left(\frac{n+2}2\right)^{n+1}}{\left(\frac{n+1}2\right)^n}&=\frac{n+2}2\left(\frac{n+2}{n+1}\right)^n\\ &=\frac{n+2}2\left(1+\frac1{n+1}\right)^n\\\\ &>\frac{n+2}2\left(1+\frac{n}{n+1}\right)\\\\ &=\frac{(n+2)(2n+1)}{2n+2}\\\\ &=\frac{2n^2+5n+2}{2n+2}\\\\ &=n+\frac{3n+2}{2n+2}\\\\ &>n+1\;. \end{align*}$$

The first inequality follows from the binomial theorem, for instance.

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We start from $$(n+1)!$$ the left side for n+1-th step. $$(n+1)!=(n+1) n!\leq (n+1) \frac{(n+1)^n}{2^n}$$ from n-th step. Rewrite: $$(n+1) \frac{(n+1)^n}{2^n}=2(n+1) \frac{(n+1)^n}{2^{n+1}}$$ Noting that $$2(n+1)^{n+1}\leq (n+2)^{n+1}$$ beause $$\left(1+\frac{1}{n+1}\right)^{n+1}\geq 2$$ ($(1+1/n)^n$ is an increasing sequence) we have that $$2(n+1) \frac{(n+1)^n}{2^{n+1}}\leq \left(\frac{n+2}{2}\right)^{n+1}$$

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