Proving inequality $\frac{4^m}{4\sqrt{m}}\le\binom{2m}{m}$ only with Chebyshev inequality

Question

What is a good approach to proving this inequality:

$$\dfrac{4^m}{4\sqrt{m}}\le\binom{2m}{m}$$

using the Chebyshev inequality: https://en.wikipedia.org/wiki/Chebyshev's_inequality

I thought about using a random variable which has a binomial distribution, but I am not sure.

Thank you!

The answer here: Elementary central binomial coefficient estimates is not good for me, since it doesn't use the Chebychev inequality, I can already do it in an elementary way.

Answer 1

I cannot quite get all the way with just Chebychev, but almost. If any others see a dumb trick I'm missing, please feel free to comment.

Let $X\sim\text{Bin}\left(2m, \displaystyle{1\over 2}\right)$. Then we have that

$$\begin{cases} P(X=n) = \displaystyle{2m\choose n}{1\over 4^m}\le {2m\choose m}{1\over 4^m} \\ \mu_X = m \\ \sigma_X = \sqrt{m\over 2} \end{cases}$$

So we can use Chebychev with $k=\sqrt{2}$ to get

$$P(|X-m|\ge\sqrt{m})\le {1\over 2}$$ $$\iff P(|X-m|<\sqrt{m})\ge {1\over 2}$$

This means

$${1\over 2}\le \sum_{n<\sqrt{m}}P(|X-m|=n)<(2\lfloor\sqrt{m}\rfloor+1){2m\choose m}{1\over 4^m}$$ $${4^m\over 4\lfloor\sqrt{m}\rfloor + 2}\le {2m\choose m}$$

which is (almost) the desired result. If we break it up a little more, then in the case $m$ is a perfect square our count is for $\displaystyle\sum_{n=0}^{\sqrt m-1}P(|X-m|=n)$ giving the better bound of

$${1\over 2}\le (2\sqrt{m}-1){2m\choose m}{1\over 4^m}.$$