Bertrand-Chebyshev Theorem/Lemma 1

Lemma
For all $n \in \N$:


 * $\dbinom {2 n} n \ge \dfrac {2^{2 n}} {2 n + 1}$

Proof
Since $\dbinom n {k + 1} = \dfrac {n - k} {k + 1} \dbinom n k$, it is evident that $\dbinom n k$ increases for $k < \dfrac n 2$ and decreases $k > \dfrac n 2$.

Therefore, $\dbinom {2 n} n$ is the largest term in the sequence $\dbinom {2 n} 0, \dbinom {2 n} 1, \ldots, \dbinom {2 n} n$.

Finally observe that the mean of the terms in the sequence is $\dfrac {2^{2 n}} {2 n + 1}$.