To prove the inequality:- $\frac{4^m}{2\sqrt{m}}\le\binom{2m}{m}\le\frac{4^m}{\sqrt{2m+1}}$

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 $$ \frac{4^n}{2\sqrt{n}}\le\binom{2n}{n}\le\frac{4^n}{\sqrt{2n+1}}\tag{3} $$

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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.