TITLE: how to prove $\sum _{|k|\lt\sqrt m}\binom{2m}{m+k}\ge2^{2m-1}$
QUESTION [6 upvotes]: how to prove $$\sum _{|k|\lt\sqrt m}\binom{2m}{m+k}\ge2^{2m-1},\forall m\ge1$$ Thanks in advance .

REPLY [7 votes]: Consider tossing a fair coin $2m$ time and let $X$ count the number of heads. Then from Chebyshev inequality, we have that
$$\mathbb{P}(\vert X  - m\vert \geq k \sqrt{m/2}) \leq \dfrac1{k^2}$$
Take $k =\sqrt{2}$, to get that
$$\mathbb{P}(\vert X  - m\vert \geq \sqrt{m}) \leq \dfrac12, \,\,\, \text{i.e., }\,\,\, \mathbb{P}(\vert X  - m\vert < \sqrt{m}) \geq \dfrac12$$
Hence, we have
$$\sum_{\vert k \vert <\sqrt{m}} \dfrac{\dbinom{2m}{m+k}}{2^{2m}} \geq \dfrac12$$
which gives us
$$\sum_{\vert k \vert <\sqrt{m}} \dbinom{2m}{m+k} \geq 2^{2m-1}$$