Solving $\frac{1}{\sqrt[n]{(2n)!}}\ge\frac{1}{n(n+1)}$ 
Solve $$\frac{1}{\sqrt[n]{(2n)!}}\ge\frac{1}{n(n+1)}$$

This is an inequality I've been attempting. It is from a past olympiad (final stage of it, was called the national olympiad, and it was from Romania). But my method of solving was incomplete or plain wrong.
I've actually tried to bring the term from the right side to the left and to do the calculations but I might have miscalculated it ( although I've triple checked). I think a miscalculation is the actual reason but that seems like the lazy solution not the smart one.
Could someone please provide maybe a smarter solution to the problem that doesn't actually involve so many useless calculations? It could be a method that’s still lazy, but I really need an answer to the question. I've been contemplating on it for quite a while now, and I can't really get to actual solving as I am currently on holiday with some friends.
Sorry for so much context. I guess that's not really that relevant but an answer would be hugely appreciated.
 A: The inequality is equivalent to
$$\iff n(n+1) \ge \sqrt[n]{(2n)!}\iff n(n+1) \ge \sqrt[n]{\prod_{k=1}^nk(2n+1-k)} \tag{1}$$
Applying the AM–GM inequality, we have
$$\sqrt[n]{\prod_{k=1}^nk(2n+1-k)} \le\frac{1}{n}\sum_{k=1}^nk(2n+1-k)=\frac{1}{n}\frac{1}{3} n (1 + n) (2n+1) = (n+1)\frac{2n+1}{3}$$
As $\frac{2n+1}{3}\le n$ for all $n \ge 1$, we conclude that $(1)$ holds true for all $n \in \Bbb N^+$
A: Here's another elementary way.
$\dfrac{1}{\sqrt[n]{(2n)!}}
\ge\dfrac{1}{n(n+1)}
$
same as
$\sqrt[n]{(2n)!}
\le n(n+1)
$
or
$(2n)!
\le (n(n+1))^n
$.
Let
$r(n)
=\dfrac{(n(n+1))^n}{(2n)!}
$.
$r(1)
=\dfrac{2}{2}
=1
$,
$r(2)
=\dfrac{6^2}{24}
\gt 1
$,
$r(3)
=\dfrac{12^3}{6!}
=\dfrac{1728}{720}
\gt 1
$.
I will show that
$r(n)$
is increasing.
$\begin{array}\\
\dfrac{r(n+1)}{r(n)}
&=\dfrac{\dfrac{((n+1)(n+2))^{n+1}}{(2n+2)!}}{\dfrac{(n(n+1))^n}{(2n)!}}\\
&=\dfrac{(n+1)(n+2)^{n+1}}{(2n+1)(2n+2)n^n}\\
&=\dfrac{(n+2)^{n+1}}{2(2n+1)n^n}\\
&=\dfrac{(n+1)(1+2/n)^{n}}{2(2n+1)}\\
&\gt\dfrac{(1+2/n)^{n}}{4}
\qquad\text{since } n+1 > n+\frac12\\
\end{array}
$
From the first 3 terms
of the binomial theorem,
$\begin{array}\\
(1+2/n)^n
&\gt 1+n(2/n)+n(n-1)(2/n)^2/2\\
&= 1+2+2(n-1)/n\\
&=5-2/n\\
&\gt 4
\qquad\text{for } n > 2\\
\end{array}
$
Therefore
$r(n+1) > r(n)$
for $n \gt 2$
so
$r(n) > 1$
for all $n$.
