<s>[INST] To prove the inequality:- $\frac{4^m}{2\sqrt{m}}\le\binom{2m}{m}\le\frac{4^m}{\sqrt{2m+1}}$ 
Problem Statment:-
Prove:-$$\dfrac{4^m}{2\sqrt{m}}\le\binom{2m}{m}\le\dfrac{4^m}{\sqrt{2m+1}}$$

My Attempt:-
We start with $\binom{2m}{m}$ (well that was obvious), to get
$$\binom{2m}{m}=\dfrac{2^m(2m-1)!!}{m!}$$
Now, since $2^m\cdot(2m-1)!!\lt2^m\cdot2^m\cdot m!\implies \dfrac{2^m\cdot(2m-1)!!}{m!}\lt 4^m$
$$\therefore \binom{2m}{m}=\dfrac{2^m(2m-1)!!}{m!}\lt4^m$$
Also, $$2^m\cdot(2m-1)!!\gt2^m\cdot(2m-2)!!\implies 2^m(2m-1)!!\gt2^m\cdot2^{m-1}\cdot(m-1)!\\
\implies \dfrac{2^m\cdot(2m-1)!!}{m!}\gt\dfrac{4^m}{2m}$$
So, all I got to was $$\dfrac{4^m}{2m}\lt\binom{2m}{m}\lt4^m$$
So, if anyone can suggest me some modifications to my proof to arrive at the final result, or just post a whole different non-induction based proof.
 [/INST] Taking the product of the ratios of the terms gives
$$
\binom{2n}{n}=\prod_{k=1}^n4\frac{k-1/2}{k}\tag{1}
$$
Bernoulli's Inequality says
$$
\sqrt{\frac{k-1}k}\le\frac{k-1/2}{k}\le\sqrt{\frac{k-1/2}{k+1/2}}\tag{2}
$$
Applying $(2)$ to $(1)$, we get
$$
\frac{4^n}{2\sqrt{n}}\le\binom{2n}{n}\le\frac{4^n}{\sqrt{2n+1}}\tag{3}
$$

In this answer, it is shown that
$$
\frac{4^n}{\sqrt{\pi\left(n+\frac13\right)}}\le\binom{2n}{n}\le\frac{4^n}{\sqrt{\pi\left(n+\frac14\right)}}\tag{4}
$$
which is a much tighter estimate.
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