<s>[INST] Prove inequality with binomial coefficient: $6 + \frac{4^n}{2 \sqrt{n}} \le \binom{2n}n$ I have to prove inequality, where $n \in N$
$$6 + \frac{4^n}{2 \sqrt{n}} \le {2n \choose n}$$
I have checked and it is true when $n\ge4$, however I have no idea how I should start. Can anyone give a hint?
 [/INST] $$\frac{1}{4^n}\binom{2n}{n} = \frac{(2n-1)!!}{(2n)!!} = \prod_{k=1}^{n}\left(1-\frac{1}{2k}\right)$$
and by squaring both sides:
$$\left(\frac{1}{4^n}\binom{2n}{n}\right)^2 = \frac{1}{4}\prod_{k=2}^{n}\left(1-\frac{1}{2k}\right)^2 = \frac{1}{4}\prod_{k=2}^{n}\left(1-\frac{1}{k}\right)\prod_{k=2}^{n}\left(1+\frac{1}{4k(k-1)}\right)$$
hence:
$$\left(\frac{1}{4^n}\binom{2n}{n}\right)^2 = \frac{1}{4n}\prod_{k=2}^{n}\left(1+\frac{1}{4k(k-1)}\right)=\frac{1}{4n}\prod_{k=2}^{n}\left(1-\frac{1}{(2k-1)^2}\right)^{-1}$$
and:
$$\left(\frac{1}{4^n}\binom{2n}{n}\right)^2 = \frac{1}{\pi n}\prod_{k>n}\left(1+\frac{1}{4k(k-1)}\right)^{-1}=\frac{1}{\pi n}\prod_{k>n}\left(1-\frac{1}{(2k-1)^2}\right)$$
but in a neighbourhood of the origin $(1+x)^{-1}\geq e^{-x}$, hence:
$$\left(\frac{1}{4^n}\binom{2n}{n}\right)^2 \geq \frac{1}{\pi n}\exp\left(-\sum_{k>n}\frac{1}{4k(k-1)}\right)=\frac{e^{-\frac{1}{4n}}}{\pi n}$$
and we have:
$$ \frac{1}{4^n}\binom{2n}{n}\geq \frac{e^{-\frac{1}{8n}}}{\sqrt{\pi n}}\geq \frac{8n-1}{8n\sqrt{\pi n}}$$
that is stronger than the given inequality for every $n\geq 5$. 
It just remains to check the first cases by hand.
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