Need to proof by induction (calculus1 n and n+1) what needs to be proven
I need to prove that for every natural n that happens and the prove needs to be by induction,
I wont prove for n=1 because its obvious,so lets move to n and n+1
$$ \frac{(2n!)}{(n!)^2} ≥ \frac{(4^n)}{(2n+1)} $$
the assumption for n+1
$$ \frac{(2n+2)!}{(n+1)!^2} ≥ \frac{(4^{n+1})}{2n+3} $$
now I dont now how to progress,I tried this for n+1
$$ \frac{(2n+2)(2n+1)(2n!)}{(n+1)!^2} ≥ \frac{(4^{n}4^1)}{(2n+1)+2)} $$
and had a suggestion to use
$ 2n! = (n!) \prod _{n=n}^{2n}\:n\cdot )! $
but cant see how it helps,ty.

$$\frac{(2n+2)!}{((n+1)!)^2}=\frac{(2n+2)(2n+1)(2n)!}{(n+1)(n+1)(n!)^2}=\frac{2(2n+1)(2n)!}{(n+1)(n!)^2}\geq\frac{4^{n+1}}{2n+3}$$
Assuming that $\dfrac{(2n)!}{(n!)^2}\geq\dfrac{4^{n}}{2n+1}$
Let's show
$$\frac{2(2n+1)}{n+1}\frac{4^n}{2n+1}>\frac{4^{n+1}}{2n+3}$$
$$\frac{2}{n+1}>\frac{4}{2n+3}$$
$$4n+6>4n+4$$
