I'm trying to prove the inequality below using induction:
$\dfrac{2^{2n}}{n+1} < \dfrac{(2n)!}{(n!)^2} \,\, (n > 1)$
The base case is ok, but I'm not able to see how I can proceed in the induction step. I'm reaching a dead end, either multiplying by 2 or adding $\dfrac {1} {k + 1}$ to both sides of the inequality. Any tips?
For the induction step we assume that for some $n$
$$\dfrac{2^{2n}}{n+1} < \dfrac{(2n)!}{(n!)^2} \tag 1$$
then
$$\dfrac{2^{2n+2}}{n+2} =\frac{2^2(n+1)}{n+2}\dfrac{2^{2n}}{n+1} \stackrel{(1)}< \frac{2^2(n+1)}{n+2}\dfrac{(2n)!}{(n!)^2}\stackrel{?}<\dfrac{(2n+2)!}{((n+1)!)^2}$$
and the latter inequality is true indeed
$$\frac{2^2(n+1)}{n+2}\dfrac{(2n)!}{(n!)^2}<\dfrac{(2n+2)!}{((n+1)!)^2}$$ $$\frac{2^2(n+1)}{n+2}<\dfrac{(2n+2)(2n+1)}{(n+1)^2}$$ $$4(n+1)^3<(2n+2)(2n+1)(n+2)$$
$$n(n+1)>0$$
