<s>[INST] Proving $\binom{2n}{n}\ge\frac{2^{2n-1}}{\sqrt{n}}$ Prove that
$$\binom{2n}{n}\ge\dfrac{2^{2n-1}}{\sqrt{n}}$$
By the way: I have see
$$\binom{2n}{n}\ge\dfrac{4^n}{2n}=\dfrac{2^{2n-1}}{n}$$
proof:
Applying the binomial theorem
$$4^n=(1+1)^{2n}=\sum_{k=0}^{2n}\binom{2n}{k}=2+\sum_{k=1}^{2n-1}\binom{2n}{k}\le 2n\binom{2n}{n}$$
becasuse
$$\binom{2n}{k}\le\binom{2n}{n},k=0,1,2,\cdots,2n$$
But $$\dfrac{2^{2n-1}}{\sqrt{n}}\ge\dfrac{2^{2n-1}}{n}$$
so my question  must use other methods?  Thank you?
 [/INST] Use the induction. Suppose it is true for some $ n $ (and it is true for $ n = 1 $):
\begin{equation}
\binom{2n}{n} \ge \frac{2^{2n-1}}{\sqrt{n}},
\end{equation}
let's have a look at $ n + 1 $:
\begin{equation}
\binom{2(n+1)}{n+1} = \binom{2n}{n}\frac{(2n+1)(2n+2)}{(n+1)(n+1)}
\end{equation}
\begin{equation}
\frac{2^{2(n+1)-1}}{\sqrt{n+1}}=\frac{2^{2n-1}}{\sqrt{n}}\frac{4\sqrt{n}}{\sqrt{n+1}}
\end{equation}
Therefore the initial statement for $n + 1$
\begin{equation}
\binom{2(n+1)}{n+1} \ge \frac{2^{2(n+1)-1}}{\sqrt{n+1}}
\end{equation}
is equivalent to 
\begin{equation}
\frac{(2n+1)(2n+2)}{(n+1)(n+1)} \ge \frac{4\sqrt{n}}{\sqrt{n+1}}
\end{equation}
\begin{equation}
\frac{(2n+1)}{(n+1)} \ge \frac{2\sqrt{n}}{\sqrt{n+1}}
\end{equation}
\begin{equation}
\frac{(2n+1)^2}{(n+1)^2} \ge \frac{4n}{n+1}
\end{equation}
\begin{equation}
(2n+1)^2 \ge 4n(n+1)
\end{equation}
\begin{equation}
1 \ge 0
\end{equation}
</s>