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?
 A: $$\begin{align}
{2(n+1)\choose n+1}&={(2n+2)!\over(n+1)!(n+1)!}\\
&={(2n+2)(2n+1)\over(n+1)(n+1)}{2n\choose n}\\
&={2(2n+1)\over(n+1)}{2n\choose n}\\
&\ge{2(2n+1)\over(n+1)}\left(6+{4^n\over2\sqrt n} \right)\quad\text{(by induction)}\\
\end{align}$$
It's quite clear that
$${2(2n+1)\over(n+1)}\cdot6\ge6$$
so it remains to show that
$${2(2n+1)\over(n+1)}\cdot{4^n\over2\sqrt n}\ge{4^{n+1}\over2\sqrt{n+1}}$$
But, after cancellations, this is equivalent to showing
$$2n+1\ge2\sqrt{n(n+1)}$$
and that's easily shown by squaring both sides.
A: $$\frac{1}{4^n}\binom{2n}{n} = \frac{(2n-1)!!}{(2n)!!} = \prod_{k=1}^{n}\left(1-\frac{1}{2k}\right)$$
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.
A: In this answer is an elementary derivation of the inequality
$$
\frac{4^n}{\sqrt{\pi(n+\frac13)}}\le\binom{2n}{n}
$$
which is stronger than the given inequality for $n\ge5$. We just need to check that the given inequality is an equality for $n=4$ and false for $n\lt4$.
