How can I prove this inequality for $n\geq 2$? How can I prove this inequality for all natural numbers $n\geq 2$?
${2n\choose n}>\frac{4^n}{n+1}$
I've tried induction but that was a dead end.
 A: For $n=2$, $${4\choose 2}=6> \frac{16}{3}$$
Assume $${{2k}\choose{k}}>\frac{4^k}{k+1}$$
Then $${{2k+2}\choose {k+1}}=\frac{(2k+2)(2k+1)}{(k+1)}{{2k}\choose k}$$
$$>\frac{(2k+2)(2k+1)}{(k+1)}\frac{4^k}{k+1}$$
$$={2(2k+1)}\frac{4^k}{k+1}$$
$$=4k\frac{4^k}{k+1}+2\frac{4^k}{k+1}$$
$$>4\frac{4^k}{k+1}$$
$$=\frac{4^{k+1}}{k+1}$$
$$>\frac{4^{k+1}}{k+2}$$
Hopefully I didnt mess up the algebra but the idea is right.
A: Induction actually does work. Let $a_n$ denote the LHS and $b_n$ denote the RHS at the $n^{\text{th}}$ step. Then
$$
\frac{a_n}{a_{n-1}} = \frac{2n(2n-1)}{n^2} = \frac{4n-2}n, \quad \frac{b_n}{b_{n-1}} = \frac{4n}{n+1}.  
$$
To figure out which factor is bigger, we consider their ratio : 
$$
\frac{\frac{4n-2}{n} }{ \frac{4n}{n+1}} = \frac{(4n-2)(n+1)}{4n^2} = \frac{4n^2+2n-2}{4n^2} \ge 1 
$$
as long as $2n-2 \ge 0$, i.e. as long as $n \ge 1$. So you can start induction at $n=2$ where $\binom{2n}n = \binom 42 = 6 > \frac{16}3 = \frac{4^2}{2+1}$. 
Hope that helps,
A: Although it might look astounding (at least to me), the following equation holds:
$$\begin{align}
\binom{2n}{n}&=\frac{(2n)!}{n!\cdot n!}=\prod_{k=1}^n\frac{2k(2k-1)}{k\cdot k}
=2^{2n}\prod_{k=1}^n\frac{2k(2k-1)}{2k\cdot2k}
=2^{2n}\prod_{k=1}^n\left(1-\frac{1}{2k}\right)
\end{align}$$
Now, for $k\ge1$, we have $2k\ge k+1$ and so $1-\frac{1}{2k}\ge1-\frac{1}{k+1}$ where we only have equality if $k=1$. Therefore, if $n\ge2$:
$$\begin{align}
\binom{2n}{n}>2^{2n}\prod_{k=1}^n\left(1-\frac{1}{k+1}\right)&=2^{2n}\prod_{k=1}^n\frac{k}{k+1}=\frac{2^{2n}}{n+1}
\end{align}$$
