So I'm trying to prove that for $\frac{1}{2}< x \leq 1$ we have
$$\sum_{\lceil nx \rceil}^{n}{n \choose k} \leq 2^{nh(x)}$$
I've managed to prove that $$\sum_{0}^{\lfloor nx \rfloor}{ n\choose k}\leq2^{nH(1-x,x)}$$
But when I try to progress from the fact that $\sum_{0}^{n}{n \choose k}=2^{n}$ I seem to be getting the inequality sign the wrong way round. Can anyone help?
$$\sum_{k=\lceil n x \rceil}^{n}{n \choose k} = \sum_{k=\lceil nx \rceil}^{n}{n \choose n-k} = \sum_{j=0}^{\lfloor n \lambda \rfloor}{n \choose j} $$
where in the later $ n - \lceil n x \rceil=\lfloor n(1-x)\rfloor = \lfloor n \lambda\rfloor $, with $\lambda = 1-x$, $0\le \lambda < 1/2$
Then see here.