Our teacher asked this question in a homework :

Prove : $\displaystyle \binom{2n}{n} \ge \frac{4^{n}}{2n+1}$

I found these pages which gave me some clues :

number theory - prove that $(2n)!/(n!)^2$ is even if $n$ is a positive integer - Mathematics

Prove that 2n choose n is less than or equal to 4^n? - Yahoo! Answers

However, I can't figure that out ! According to the 2nd page, if the right member was only $\displaystyle 4^n$, then the left member would be less than or equal to the right member.

I just don't know what to do.

You help would be appreciated.

Thanks.