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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?

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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}

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nice!+1, Thank you very much – math110 Aug 11 '13 at 3:52

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