Is it true that for $n \ge5$, ${{3n} \choose {2n}} > \frac{6^n}{n}$ For $n \ge 5, \frac{9n^2 - 9n + 2}{4n^2-2n} > 2$

$9n^2 -9n +2 -8n^2+4n = n^2 -5n +2 \ge 5^2-25+2$

Here's the basis step:

$${{9} \choose {6}} = 84 > \frac{6^3}{3} = 72$$

Here's the inductive step:

$${{3n} \choose {2n}} = {{3(n-1)}\choose {2(n-1)}}\left(\frac{3n}{n}\right)\left(\frac{3n-1}{2n-1}\right)\left(\frac{3n-2}{2n}\right) > \left(\frac{6^{n-1}}{n-1}\right)\left(\frac{3n}{n}\right)\left(\frac{9n^2 -9n +2}{4n^2-2n}\right) > \left(\frac{6^{n-1}}{n}\right)(6) = \frac{6^n}{n}$$

 A: $$\begin{eqnarray*}\log\binom{3n}{2n} &=& \log\Gamma(3n+1)-\log\Gamma(n+1)-\log\Gamma(2n+1)\\ &=& -\log n-\log B(n,2n+1)\end{eqnarray*}$$
and
$$ B(n,2n+1) = \int_{0}^{1}\left[x(1-x)^2\right]^n\,\frac{dx}{x} $$
is a log-convex function by the Cauchy-Schwarz inequality.
You don't even need induction.
A: I will show that
$0.850
\lt \dfrac{\binom{3n}{2n}}{\sqrt{\dfrac{3}{4\pi n}}
\left(\dfrac{27}{4}\right)^n}
\lt 1.085
$.
More precise bounds,
of the form
$1+O(\frac{c}{n})$,
can be easily gotten
by the method below.
Since
$n! \approx \sqrt{2\pi n}(n/e)^n$,
$\begin{array}\\
\binom{an}{bn}
&=\dfrac{(an)!}{(bn)!((a-b)n!}\\
&\sim \dfrac{\sqrt{2\pi an}(an/e)^{an}}
{(\sqrt{2\pi bn}(bn/e)^{bn})(\sqrt{2\pi n(a-b)}(((a-b)n)/e)^{(a-b)n})}\\
&= \sqrt{\dfrac{2\pi an}{2\pi bn2\pi n(a-b)}}\left(\dfrac{(an)^ae^be^{a-b}}{e^a(bn)^b((a-b)n)^{a-b}}\right)^n\\
&= \sqrt{\dfrac{ a}{2\pi bn(a-b)}}\left(\dfrac{a^an^a}{b^b(a-b)^{a-b}n^a}\right)^n\\
&= \sqrt{\dfrac{ a}{2\pi bn(a-b)}}\left(\dfrac{a^a}{b^b(a-b)^{a-b}}\right)^n\\
&=b(n, a, b)\\
\end{array}
$
If $a=3, b=2$,
$\dfrac{a}{b(a-b)}
=\dfrac{3}{2}
$
and
$\dfrac{a^a}{b^b(a-b)^{a-b}}
=\dfrac{3^3}{2^2}
=\dfrac{27}{4}
=6\frac34
$
so
$\binom{3n}{2n}
\sim \sqrt{\dfrac{3}{4\pi n}}
\left(\dfrac{27}{4}\right)^n
$.
Actually
 (https://en.wikipedia.org/wiki/Stirling%27s_approximation),
 if
 $f(n)
 =\dfrac{n!}{\sqrt{2\pi n}(n/e)^n}$,
 then the following inequality holds:
$$1
 \lt f(n)
 \lt  e^{1/(12n)}
 \lt \frac{e}{\sqrt{2\pi}} 
 = 1.0844...
 $$
So,
if $u < f(n) < v$,
then
$\frac{u}{v^2}
\lt \dfrac{\binom{an}{bn}}{b(n, a, b)}
\lt \frac{v}{u^2}
$.
From the simple bounds above,
we can take
$u = 1$
and
$v = \frac{e}{\sqrt{2\pi}} $,
so the bounds are
$\dfrac{2\pi}{e^2}
=0.850...
$
and
$\dfrac{e}{\sqrt{2\pi}}
=1.0844...
$.
Therefore
$0.850
\lt \dfrac{\binom{3n}{2n}}{\sqrt{\dfrac{3}{4\pi n}}
\left(\dfrac{27}{4}\right)^n}
\lt 1.085
$.
