$4^{n-1} {{3n}\choose{n}} \sqrt{n} \geq 3^{3n-2}, n \in \mathbb{N} \setminus \{0\}$ Induction proof, need a hint for more efective proof. I am looking for a smarter way of proving that inequality by induction.
$$4^{n-1} {{3n}\choose{n}} \sqrt{n} \geq 3^{3n-2}, n \in \mathbb{N} \setminus \{0\}$$
For $n + 1$:
$$4^{n} {{3n + 3}\choose{n+1}} \sqrt{n+1} \geq 3^{3n+1}$$
$$4*4^{n-1} \frac{(3n)!(3n + 1)(3n + 2)(3n + 3)}{(n)!(n+1)(2n)!(2n+1)(2n+2)} \sqrt{n+1} \geq 3^{3n+1}$$
Then I can plug my thesis into the inequality for $n + 1$:
$$4^{n-1} {{3n}\choose{n}} \sqrt{n} \geq 3^{3n-2} \implies 4^{n-1} \frac{(3n)!}{n!(2n)!} \geq \frac{3^{3n-2}}{\sqrt{n}}$$
So I get:
$$4* \frac{3^{3n-2}}{\sqrt{n}} * \frac{(3n + 1)(3n + 2)(3n + 3)}{(n+1)(2n+1)(2n+2)} \sqrt{n+1} \geq 3^{3n+1}$$
$$4*(3n + 1)(3n + 2)(3n + 3) \sqrt{n+1} \geq 27\sqrt{n} (n+1)(2n+1)(2n+2) $$
And then, if I sqare both hand sides of that ineqality (both are positive) I get 7th powers of $n$ and there is probably some smarter / easier way to prove that inequality.

This isn't induction.
It uses Stirling
and it is more general.
I will show that
$0.850
\lt \dfrac{\binom{3n}{n}}{\sqrt{\dfrac{3}{4\pi n}}
\left(\dfrac{27}{4}\right)^n}
\lt 1.085
$.
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\\
&=r(n, a, b)\\
\end{array}
$
If $a=3, b=1$,
$\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}{n}
\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}}{r(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{an}{bn}}{\sqrt{\dfrac{ a}{2\pi bn(a-b)}}\left(\dfrac{a^a}{b^b(a-b)^{a-b}}\right)^n}
\lt 1.085
$.
If $a=3, b=1$ then
$0.850
\lt \dfrac{\binom{3n}{n}}{\sqrt{\dfrac{3}{4\pi n}}
\left(\dfrac{27}{4}\right)^n}
\lt 1.085
$.
