Proving ${(n+2)!}< \left(n(\sqrt{2}-1)+\sqrt{2}\right)^{n+2}$ While simplifying an inequality, this inequality was derived:
$${(n+2)!}< \left(n(\sqrt{2}-1)+\sqrt{2}\right)^{n+2},\quad\quad\quad\quad n\in \mathbb{N}$$
Do you have any idea to prove it? It is very close for small $n$.
 A: Useful bounds on $n!$ (rather than asymptotics) are given here; in this case, you can use the fact that
$$
n! \le e n^{n+1/2}e^{-n}
$$
for all $n\ge 1$.  Taking the $n$-th root of both sides,
$$
\sqrt[n]{n!} \le \left(\frac{n}{e}\right)\left(e \sqrt{n}\right)^{1/n}.
$$
Now,
$$
\left(e\sqrt{n}\right)^{1/n}=\exp\left({\frac{1}{n}\ln (e\sqrt{n})}\right)=\exp\left({\frac{2+\ln n}{2n}}\right) \rightarrow 1,
$$
decreasing monotonically from above; in particular it's less than $9/8$ for $n\ge 22$ (by a direct calculation).
You want to show that
$$
\sqrt[n]{n!}<(\sqrt{2}-1)(n-2)+\sqrt{2}=(\sqrt{2}-1)n+(2-\sqrt{2})
$$
for all $n\ge 3$.  The above calculation shows the stronger inequality
$$
\sqrt[n]{n!}<\frac{9}{8e}n<(\sqrt{2}-1)n$$
for $n \ge 22$.  The remaining cases, $n=3,4,\ldots,21,$ can then be shown by hand to satisfy the original inequality.  As you point out, for $n=3$ and $n=4$ it is close (and of course for $n=1$ and $n=2$ you have equality).
A: Using Stirling,
$$n!\approx n^ne^{-n}\sqrt {2\pi n}$$
we can approximate the left hand side with
$$ (n+2)!\approx (n+2)^{n+2}e^{-n-2}\sqrt{2\pi (n+2)}$$
and the right hand side with
$$ (n(\sqrt 2-1)+\sqrt2)^{n+2}>n^{n+2}(\sqrt2-1)^{n+2}=(n+2)^{n+2}\left(1+\frac2n\right)^{-(n+2)}(\sqrt2-1)^{n+2}.$$
If we drop the common factor $(n+2)^{n+2}$ and note that $\left(1+\frac2n\right)^{-(n+2)}=\left(1-\frac2{n+1}\right)^{n+2}\to e^{-2}$, we need only compare the growth of 
$\sqrt{2\pi (n+2)}$ and $((\sqrt 2-1)e)^{n+2}$ . Since $(\sqrt 2-1)e>1$, the exponential "wins" for $n$ big enough. To turn this into a proper proof, you need to take a precise look at the error hidden in the "$\approx$" of Stirling and tthus determine, what "$n$ big enough" means. It turns out that the relative error is below $\frac 1{12n}$ and therefore $n=22$ can be taken as "big enough".
EDIT: Thanks to CutieKrait's comment, I noted that I had $(1+\frac2n)^{n+2}$ instead of its reciprocal. Thus one needs to have a closer look at its value as well (it's simply no longer "$>1$", but at least bounded from below ...)
A: A completely different approach, that's why I don't simply edit my first (lousy) attempt:
We want to show that
$$\tag1 n!<(\alpha n+\beta)^n\quad\text{for }n\ge 3,$$
where $\alpha=\sqrt 2-1\approx0.414$ and $\beta=\sqrt 2-2\alpha=2-\sqrt 2\approx0.586$.
Assume we findan inequality 
$$ \tag2 n!\le (\gamma n)^n\quad\text{for }n\ge n_0.$$
Then we have shown $(1)$ for all $n$ with $n_0\le n\le \frac\beta{\gamma-\alpha}$  if $\gamma>\alpha$. Of course, if $(2)$ holds even for some $\gamma <\alpha$, then we have shown $(1)$ for all $n\ge n_0$.
We can try to find good such $\gamma$.
By the AGM inequality, we have 
$$\prod_{k=r}^s k\le \left(\frac{r+s}2\right)^{s-r+1}.$$
With $r=1$ this gives us immediatyly $(2)$ with $\gamma=\frac12$, hence $(1)$ for $n\le 6$ because $\frac\beta{\frac12-\alpha}\approx6.8$.
Let $a$ be a number with $0<a<1$ such that $an$ is an integer. Then
$$\tag3n!=\prod_{k=1}^{an-1}k\cdot \prod_{k=an}^nk\le \left(\frac{an}{2}\right)^{an}\left(\frac{(1+a)n}{2}\right)^{n-an}=\left(\frac n2a^a(1+a)^{1-a}\right)^n.$$
Let $\gamma=\frac{a^a(1+a)^{1-a}}2$.
Numerically, if $0.19<a<0.19+\frac17$ (which can definitely be achieved if $n\ge 7$), then $\gamma<0.42$, which gives us $(1)$ for $n\le 101$ (because $\frac{\beta}{0.42-\alpha}\approx101.2$).
These results alone may already be helpful for other methods that establish $(1)$ for $n$ "big enough". However, we could not possibly proceed significantly further up.
We can refine $(3)$: If $an-1\ge 7$, we have already seen that we can estimate $(an-1)!\le (0.42an)^{an}$ and obtain
$$\tag4 n!\le (0.42an)^{an}\left(\frac{(1+a)n}2\right)^{n-an}=(\gamma n)^n $$
with
$$\gamma = (0.42a)^a\left(\frac{1+a}2\right)^{1-a}.$$
If $0.3<a<0.6$ we find numerically that $\gamma<0.41<\alpha$.
This shows $(1)$ for all $n\ge 27$ (so that $an-1\ge 7$).
In summary, $(1)$ holds for all $n\ge 3$.
Remark: Note that a bit of work is hidden in what appears above a "numerical findings": You need to show that the expressions for $\gamma$ have a unique local minimum for $a\in(0,1)$, which then allows us to conclude $\gamma(a)\le\max\{\gamma(x_1),\gamma(x_2)\}$ for all $a\in[x_1,x_2]$.
Edit: It seems like I have made an off-by-one error somewhere, but don't see where exactly right now ...
A: Let 
$$a_k=\sqrt[k] {k!}$$
By Prove elementarily that $\sqrt[n+1] {(n+1)!} - \sqrt[n] {n!}$ is strictly decreasing
we have
$$a_{k+2}-a_{k+1}< a_{k+1}-a_k$$
taking sum we get this inequality:
$$\sum_{k=1}^{m}{a_{k+2}-a_{k+1}}< \sum_{k=1}^{m}{a_{k+1}-a_k}$$
that is
$$a_{m+2}-a_{2}< a_{m+1}-a_1$$
that is
$$a_{m+2}-a_{m+1}< a_{2}-a_1=\sqrt{2}-1$$
taking sum again we get this inequality:
$$\sum_{m=1}^{n}{a_{m+2}-a_{m+1}}< n(\sqrt{2}-1)$$
that is:
$$a_{n+2}-a_{2}<n(\sqrt{2}-1)$$
or
$$a_{n+2}<n(\sqrt{2}-1)+\sqrt{2}$$
or
$$\sqrt[n+2] {(n+2)!}< n(\sqrt{2}-1)+\sqrt{2}$$
or
$${(n+2)!}< \left(n(\sqrt{2}-1)+\sqrt{2}\right)^{n+2}$$
