Some trouble with the induction 
Prove, that for any positive integer $n \geqslant 2$ we have the
  inequality $$ \frac{ 4^n }{ n+1 } < \frac{ (2n)! }{ (n!)^2 }.$$


For $n=2$ the inequality is true. Directly just take and prove inequality for $k+1$ problematically. So, I think we need to find the recurrence relation to one of the members of inequality. Need some hint!

trivial induction:
base case $n=2$ holds.  assume ${\frac{4^n}{n+1}}\lt {\frac{(2n)!}{(n!)^2}},n\ge 2$ 
${\frac{4^n}{n+1}}(\frac{4(n+1)}{n+2})\lt {\frac{(2n)!}{(n!)^2}}(\frac{(2n+2)(2n+1)}{(n+1)(n+1)})\space$?
${\frac{4^n}{n+1}}(\frac{4n+4}{n+2})\lt {\frac{(2n)!}{(n!)^2}}(\frac{2(n+1)(2n+1)}{(n+1)(n+1)})\space$?
${\frac{4^n}{n+1}}(2)\lt {\frac{(2n)!}{(n!)^2}}(\frac{2(2n+1)}{(n+1)})\space$?
${\frac{4^n}{n+1}}(2)\lt {\frac{(2n)!}{(n!)^2}}(\frac{4n+2}{(n+1)})\space$?
${\frac{4^n}{n+1}}(2)\lt {\frac{(2n)!}{(n!)^2}}(\frac{2n+2n+2}{(n+1)})\space$?
${\frac{4^n}{n+1}}(2)\lt {\frac{(2n)!}{(n!)^2}}({\frac{2n}{(n+1)}}+{\frac{2n+2}{(n+1)}})\space$?
($? = $not taken as true just there for simplification)
${\frac{4^n}{n+1}}(2)\lt {\frac{(2n)!}{(n!)^2}}({\frac{2n}{(n+1)}}+2),n\ge 2$
obvious that  $\space 2\lt {\frac{2n}{(n+1)}}+2$ for all $n\in N$
$\therefore {\frac{4^{n+1}}{(n+1)+1}}\lt {\frac{(2(n+1))!}{((n+1)!)^2}},n\ge 2$
and it holds by Induction
