Proof of inequality $n!\leq(\frac{n+1}{2})^n$

Question

I need to prove the following inequality:

$n!\leq(\frac{n+1}{2})^n$

Thank you very much for your help.

Answer 1

Hint: Prove that the function $\log(x)$ is concave (Wikipedia). Conclude that $$\log(1)+\log(2)+\cdots+\log(n)\leq n\log(\tfrac{n+1}{2})$$

Answer 2

By AM-GM inequality

$$ \frac{1+2+3+\ldots+n}{n} \geq (1\cdot 2\cdot3\cdots n)^{1/n}$$

Hence

$$\left(\frac{n+1}{2}\right)^n \geq n!$$

Answer 3

$$[n!]^2=[n!\cdot n!=]=[1\cdot 2\cdot 3\cdots (n-2)\cdot (n-1)\cdot n\cdot n\cdot(n-1)\cdots 2\cdot 1]=$$ $$=[1\cdot n]\cdot[2\cdot (n-1)]\cdots[n\cdot 1]\leq\left(\frac{n+1}{2}\right)^2\cdots\left(\frac{n+1}{2}\right)^2=\left(\frac{n+1}{2}\right)^{2n}$$ $\Rightarrow$ $$n!\leq\left(\frac{n+1}{2}\right)^{n}$$

Answer 4

We know that $(a-b)(a+b)\le a^2$. Using the same knowledge here:

$1(n) < 2(n-1) < 3(n-2) ....< (\frac{n}{2})( \frac{n+2}{2}) < (\frac{n+1}{2})^2$

By multiplying all the terms, we obtain the result $n! \le (\frac{n+1}{2})^n$