# 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?

-

Use the induction. Suppose it is true for some $n$ (and it is true for $n = 1$):
$$\binom{2n}{n} \ge \frac{2^{2n-1}}{\sqrt{n}},$$
let's have a look at $n + 1$:
$$\binom{2(n+1)}{n+1} = \binom{2n}{n}\frac{(2n+1)(2n+2)}{(n+1)(n+1)}$$ $$\frac{2^{2(n+1)-1}}{\sqrt{n+1}}=\frac{2^{2n-1}}{\sqrt{n}}\frac{4\sqrt{n}}{\sqrt{n+1}}$$ Therefore the initial statement for $n + 1$ $$\binom{2(n+1)}{n+1} \ge \frac{2^{2(n+1)-1}}{\sqrt{n+1}}$$ is equivalent to $$\frac{(2n+1)(2n+2)}{(n+1)(n+1)} \ge \frac{4\sqrt{n}}{\sqrt{n+1}}$$ $$\frac{(2n+1)}{(n+1)} \ge \frac{2\sqrt{n}}{\sqrt{n+1}}$$ $$\frac{(2n+1)^2}{(n+1)^2} \ge \frac{4n}{n+1}$$ $$(2n+1)^2 \ge 4n(n+1)$$ $$1 \ge 0$$