Is it true that for $n \ge5$, ${{3n} \choose {2n}} > \frac{6^n}{n}$

Question

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:

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

Answer 1

$$\begin{eqnarray*}\log\binom{3n}{2n} &=& \log\Gamma(3n+1)-\log\Gamma(n+1)-\log\Gamma(2n+1)\\ &=& -\log n-\log B(n,2n+1)\end{eqnarray*}$$ 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.

Answer 2

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 $.