I encountered the following claim
$$\frac{1}{n+1}2^{nH_2(k/n)} \le \binom{n}{k} \le 2^{nH_2(k/n)}$$
where $H_2$ is the binary entropy function. The upper bound is rather well known but how does one show the lower bound?
Inequality – Lower Bound on Binomial Coefficient
binomial-coefficientsinequalityinformation theory
Best Answer
The lower bound is a rewriting of $\int_0^1 x^k (1-x)^{n-k} \leq 2^{-nH_2(k/n)}$, which is estimation of the integral by (maximum value of function integrated, which occurs at $x=\frac{k}{n}$) x (length of interval).