# 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?

\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.

• +1: Very elementary! This is helped by the fact that the given inequality is not very close at all for large $n$.
– robjohn
Commented Oct 29, 2015 at 22:06

$$\frac{1}{4^n}\binom{2n}{n} = \frac{(2n-1)!!}{(2n)!!} = \prod_{k=1}^{n}\left(1-\frac{1}{2k}\right)$$ 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.

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 $n\lt4$.