TITLE: Prove inequality with binomial coefficient: $6 + \frac{4^n}{2 \sqrt{n}} \le \binom{2n}n$
QUESTION [8 upvotes]: 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?

REPLY [7 votes]: $$\begin{align}
{2(n+1)\choose n+1}&={(2n+2)!\over(n+1)!(n+1)!}\\
&={(2n+2)(2n+1)\over(n+1)(n+1)}{2n\choose n}\\
&={2(2n+1)\over(n+1)}{2n\choose n}\\
&\ge{2(2n+1)\over(n+1)}\left(6+{4^n\over2\sqrt n} \right)\quad\text{(by induction)}\\
\end{align}$$
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.