Show that $\frac{2^{2n}}{2n} \le {2n \choose n}$, $n\in\mathbb{N}$.
What I can see easily is that $\frac{2^{2n}}{2n + 1} \le {2n \choose n}$, since $(1+1)^{2n} = \sum_{m=0}^{2n}{2n \choose m} = 2^{2n}$. This sum contains $2n+1$ terms with the largest being ${2n \choose n}$. So ${2n \choose n}$ is larger than $\frac{2^{2n}}{2n + 1}$. However I have found the statement $\frac{2^{2n}}{2n} \le {2n \choose n}$, which is not obvious to me.
The first and last terms are $\left( \begin{array}{c} 2n \\ 0 \end{array} \right) = \left( \begin{array}{c} 2n \\ 2n \end{array} \right) = 1$.
So with $n \ge 1$, the sum of these two terms is $\le \left( \begin{array}{c} 2n \\ n \end{array} \right)$ and we only have $2n$ terms in the sum now.
You can also prove it by induction on $n$; note that
$$\frac{\frac{2^{2n+2}}{2n+2}}{\frac{2^{2n}}{2n}}=\frac{4n}{n+1}<\frac{2(2n+1)}{n+1}=\frac{\binom{2n+2}{n+1}}{\binom{2n}n}\;.$$
