Bertrand-Chebyshev Theorem/Lemma 1

Lemma

For all $n \in \N$:

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

where $\dbinom {2 n} n$ denotes a binomial coefficient.

Proof

From Sequence of Binomial Coefficients is Strictly Increasing to Half Upper Index, $\dbinom n k$ increases for $k < \dfrac n 2$.

From Sequence of Binomial Coefficients is Strictly Decreasing from Half Upper Index, $\dbinom n k$ decreases for $k > \dfrac n 2$.

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

Finally, observe that the mean of the terms in the sequence is $\dfrac {2^{2 n}} {2 n + 1}$, by Sum of Binomial Coefficients over Lower Index.

$\Box$