For $n \ge 0$, is $\dfrac{\left(n+\left\lfloor\frac{2n}{3}\right\rfloor\right)!}{n!} > \dfrac{(2n)!}{\left(2n-\left\lfloor\frac{n}{3}\right\rfloor\right)!}$
I will consider an answer using the gamma function! I have used the floor function in my question only because I was not clear how to make the argument using the gamma function.
I believe the answer is yes. Here's my argument:
For all $n$, $\left\lfloor\frac{2n}{3}\right\rfloor \ge 2\left\lfloor\frac{n}{3}\right\rfloor$
So, there are $2$ terms $(n+1)(n+2)\dots(n+\frac{2n}{3})$ for every $1$ term of $(2n)(2n-1)\dots(2n-\frac{n}{3}+1)$ and since $(n+1)(n+2)=n^2+3n+2 > 2n$ where $(n+1)*(n+2)$ is the least product and $2n$ is the greatest value.
Edit: I am open to an answer with the gamma function.
We are stating that $$ 2\log\Gamma\left(\frac{5}{3}n+1\right)>\log\Gamma(2n+1)+\log\Gamma(n+1)\tag{1}$$ holds for any $n\geq 0$. In order to prove it is enough to show that
$$ 2\sum_{m\geq 1}\left(\frac{1}{m}-\frac{1}{m+\frac{5}{3}n}\right)> \sum_{m\geq 1}\left(\frac{1}{m}-\frac{1}{m+2n}\right)+\sum_{m\geq 1}\left(\frac{1}{m}-\frac{1}{m+n}\right)\tag{2}$$ then integrate both sides of $(2)$ with respect to the $n$ variable. $(2)$ is equivalent to $$ \sum_{m\geq 1}\left(\frac{1}{m+2n}+\frac{1}{m+n}-\frac{2}{m+\frac{5}{3}n}\right)>0 \tag{3} $$ but that is trivial since the main term of $(3)$ can be represented as $\frac{n (m+3 n)}{(m+n) (m+2 n) (3 m+5 n)}>0$.