Bertrand-Chebyshev Theorem/Lemma 1
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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.
$\Box$