Why is $(n+1)^{n-1}(n+2)^n>3^n(n!)^2$ 
Why is $(n+1)^{n-1}(n+2)^n>3^n(n!)^2$ for $n>1$

I can use $$(n+1)^n>(2n)!!=n!2^n$$ but in the my case, the exponent is always decreased by $1$, for the moment I don't care about it, I apply the same for $n+2$
$(n+2)^{n+1}>(2n+2)!!=(n+1)!2^{n+1}$  
gathering everything together,
$(n+1)^{n-1}(n+2)^n=\frac{(n+1)^n(n+2)^{n+1}}{(n+1)(n+2)}>\frac{(n+1)(n!)^22^{2n+1}}{(n+1)(n+2)}$
$\iff(n+1)^{n-1}(n+2)^n>(n!)^2\times\frac{2^{2n+1}}{(n+2)}$
but $\frac{2^{2n+1}}{(n+2)}>3^n$ is not true for $n=2$
can you suggest another approach ?
 A: Hint
Induction for the step $n+1$:
$$(n+2)^{n}(n+3)^{n+1}=\frac{(n+3)^{n+1}}{(n+1)^{n-1}}(n+1)^{n-1}(n+2)^{n}>3^n(n!)^2\frac{(n+3)^{n+1}}{(n+1)^{n-1}}$$
We may expect
$$\frac{(n+3)^{n+1}}{(n+1)^{n-1}}>3(n+1)^2 \quad (1)$$
in order to finish the induction.
Backing to $(1)$ we have an equivalent expression:
$$\left(\frac{n+3}{n+1}\right)^{n+1}>3 \Leftrightarrow \left(1+\frac{2}{n+1}\right)^{n+1}>3$$
Using the Bernoulli inequality $(1+x)^m \ge1+mx$ for $x>-1$.
Taking $m=n+1$ and $x=\frac{2}{n+1}$ we get what we want.
A: prove: $(n+1)^{n−1}(n+2)^n>3^n(n!)^2=3^n(n!n!)$ for $n>1$
$n=2:3^14^2=48>3^2(2)(2)=36$
assume: $(n+1)^{n−1}(n+2)^n>3^n(n!n!)$
need to arrive at: $(n+2)^{n}(n+3)^{n+1}>3^{n+1}(n+1)!(n+1)!$
for lhs need to multiply by: 
${{(n+2)^{n}(n+3)^{n+1}}\over {(n+1)^{n−1}(n+2)^n}}
={{(n+3)^n(n+3)(n+1)}\over {(n+1)^n}}=({{n+3}\over {n+1}})^n(n+3)(n+1)$
where $({{n+3}\over {n+1}})^n$ is greater than $3$ for $n=3$ and increasing for all $n\in N$. 
for rhs need to multiply by: $3(n+1)(n+1)$.  therefore lhs > rhs and 
$(n+1)^{n−1}(n+2)^n>3^n(n!)^2$ for $n>1$ by mathematical induction.
A: Let's see if
something simple works.
You want
$(n+1)^{n-1}(n+2)^n
>3^n(n!)^2
$.
First,
$(n+1)^{n-1}(n+2)^n
> n^{2n-1}
$.
Second,
since
$(n/e)^n < n! < (n/e)^{n+1}
$
(easily proved by induction from
$(1+1/n)^n < e < (1+1/n)^{n+1}$),
$3^n (n!)^2
< 3^n(n/e)^{n+1}
= (3n/e)^n(n/e)
$.
Therefore,
if
$n^{2n-1}
> (3n/e)^n(n/e)
$,
we are done.
This is the same as
$(ne/3)^n > n^2/e
$
or
$ne/3 
> (n^2/e)^{1/n}
= (n^{1/n})^2/e^{1/n}
$.
But
$n^{1/n}
< e^{1/e}
<1.5
$
and
$e^{1/n} > 1$
so
$(n^{1/n})^2/e^{1/n}
< 1.5^2
= 2.25
$.
Therefore
$ne/3 
> (n^{1/n})^2/e^{1/n}
$
if
$ne/3 > 2.25$
or
$n > 3\cdot 2.25/ e
\approx 2.48
$.
Smaller $n$ easily verified.

Someone commented "How do you know that
$e^{1/e} < 1.5$?". My answer was "Calculator."
Here is a calculator-free answer.
But
$n^{1/n}
< e^{1/e}
< \sqrt{3}
$
(since
$2 <e < 3$)
and
$e^{1/n} > 1$
so
$(n^{1/n})^2/e^{1/n}
< (\sqrt{3})^2
= 3
$.
Therefore
$ne/3 
> (n^{1/n})^2/e^{1/n}
$
if
$ne/3 > 3$
or
$n > 3\cdot 3/ e
= 9/e
\approx 3.31
$.
Smaller $n$ easily verified.
A: EDIT: This answer is wrong, because I mixed up my left and right-hand sides right at the end. I think it is salvageable, but it'll be quite a bit of work.
I'll do it without induction.
Rearrange: we want $\left(\frac{(n+1)(n+2)}{3}\right)^{n-1} \frac{n+2}{3} > (n!)^2$
We'll show that this actually holds if we remove the $\frac{n+2}{3}$ term (which is always $\geq 1$ anyway).
The right-hand side of the modified desired inequality is is $$2^2 \times 3^2 \times \dots \times n^2$$ with $n-1$ terms.
The left-hand side is $$\frac{(n+1)(n+2)}{3} \times \dots \times \frac{(n+1)(n+2)}{3}$$ with $n-1$ terms again.
But $\frac{(n+1)(n+2)}{3}$ is bigger than or equal to the $i^2$ term whenever $i$ is less than or equal to $\sqrt{\frac{(n+1)(n+2)}{3}}$, and that's $\leq n$ whenever $n > 2$. So if $n>2$, for every $i \leq n$ we have each right-hand term less than its corresponding left-hand term.
So we only need to check $n=1$ and $n=2$, and they're very easy.
