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

Question

I don't know what to do with this. Nothing works. I hope somebody can help me to find a decision $$n!\leq {\left(\frac{n+1}{2}\right)}^{n}$$

Answer 1

Write, for $n\geq1$,

$$n!=1\times2\times\dots\times n\\=n\times (n-1)\times\dots\times 1$$

Then, multiply corresponding factors:

$$n!^2=(n\times1)\times((n-1)\times2)\times\dots\times(1\times n)$$

Inside the paretheses, you have $i\times(n+1-i)$ for $i\in\{1\dots n\}$.

Next, notice that $i\times (n+1-i)$, as a function of real $i$, is maximum for $\frac{n+1}2$ (it's a parabola, and the maximum is obtained in the midpoint between the roots). And this maximum is $\frac{n+1}{2}\left(n+1-\frac{n+1}{2}\right)=\left(\frac{n+1}{2}\right)^2$.

So each parenthesis in the preceding product is less than or equal to $\left(\frac{n+1}{2}\right)^2$, and

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

Finally

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

Answer 2

The claim is a straightforward consequence of the AM-GM inequality, $$ n!=\prod_{k=1}^{n}k<\left(\frac{1}{n}\sum_{k=1}^{n}k\right)^n = \left(\frac{n+1}{2}\right)^n.$$ For an improved bound, we may consider that for any $n\geq 2$ $$ \log(n!)=\sum_{k=1}^{n}\log(k) = n\log(n)-\sum_{k=1}^{n-1}k \log\left(1+\frac{1}{k}\right) $$ follows from summation by parts, and since $\frac{\log(1+x)}{x}>\frac{1}{1+\frac{x}{2}}$ over the interval $(0,1)$, $$ \log(n!) < n(\log n-1)+\sum_{k=0}^{n-1}\frac{1}{2k+1}<n(\log n-1)+1+\frac{\log n}{2} $$ hence, by exponentiation, $$ n! < \color{red}{e\sqrt{n}\left(\frac{n}{e}\right)^n}.$$

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By refining the last approach through creative telescoping, we may also prove Stirling's inequality: $$ \left(\frac{n}{e}\right)^n \sqrt{2\pi n} \exp\left(\frac{1}{12n+1}\right) < n! < \left(\frac{n}{e}\right)^n \sqrt{2\pi n} \exp\left(\frac{1}{12n}\right).$$

Answer 3

Proof by induction.

$$1! \le \dfrac{(1+1)\ ^ 1}{2} = 1 $$ Assume for all n (that the inequality holds) , and prove for n+1: $$ (n+1)! = (n+1)*n! \le (n+1)*\dfrac{(n+1) \ ^ n}{2} = \dfrac{(n+1) \ ^ {n+1}}{2} \le \dfrac{(n+1 +1) \ ^ {n+1}}{2} = \dfrac{(n+2) \ ^ {n+1}}{2}$$

in the first inequality I used the I.H.

Answer 4

Induction is a a nice way to go here, but also we can do a direct proof by grouping the numbers.

First of all note that $(n+1-k)k \le \left(\frac{n+1}{2}\right)^2$, for $1 \le k \le n$ which follows by expanding the inequality.

Now just multiply all such inequalities and you will get the wanted inequality.

Answer 5

$$\begin{align} n!\le\left(n+1\over2\right)^n&\iff\ln1+\ln2+\cdots+\ln n\le\ln\left(n+1\over2\right)+\ln\left(n+1\over2\right)+\cdots+\ln\left(n+1\over2\right)\\ &\iff\ln\left(2\over n+1\right)+\ln\left(4\over n+1\right)+\cdots+\ln\left(2n\over n+1\right)\le0\\ &\iff\ln\left(2\cdot2n\over (n+1)^2\right)+\ln\left(4(2n-2)\over (n+1)^2\right)+\cdots+\ln\left(2n\cdot2\over (n+1)^2\right)\le0\\ \end{align}$$

where the left hand side in last line comes from doubling the line above it and pairing the $k$th term with the $(n+1-k)$th term. It remains to show that

$${2k(2n+2-2k)\over(n+1)^2}\le1\quad\text{for }1\le k\le n$$

Writing $N$ for $n+1$, we have

$${2k(2N-2k)\over N^2}\le1\iff0\le N^2-4kN+4k^2=(N-2k)^2$$