I am trying to find the smallest $0< \delta < 1/2$ so that the following bound holds:
$$\sum_{k=0}^{\delta n} {n \choose k}\left(\sum_{i=k}^{n}{n \choose i}\right)\leq \frac{2^{2n}}{4n} $$
Plotting this, it seems like $\delta \leq (\frac{1}{2}- \frac{c}{\sqrt{n}})$ is plenty when $c \geq 2$ (I think the actual value of $c$ probably resembles the constant in Sterling's formula). I'm having trouble actually proving this however.
I tried upper bounding using the Entropy based bound for sums of binomial coefficients:
$$\frac{2^{h(\delta)\cdot n}}{n} \leq \sum_{k=0}^{\delta n}{n \choose k} \leq 2^{h(\delta)\cdot n}$$
(where $h(\cdot)$ is the binary entropy function) which holds for $\delta<\frac{1}{2}$, But this bound doesn't seem to be tight enough. Considering out-of-the-box bounds for binary entropy, this approach only gives $\delta \leq (\frac{1}{2}- \frac{c}{\sqrt{n}})$ for $c= \omega(1)$. Ideally, I would like to show that this holds for $c=O(1)$ suffices. Any help would be greatly appreciated.