Question: > 
> Prove that $$\binom{2n}{n}>\frac{4n}{n+1}\forall \; n\geq 2, n\in \mathbb{N}$$
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$\bf{My\; Try::}$ Using $$(1+x)^n = \binom{n}{0}+\binom{n}{1}x+\binom{n}{2}x^2+........+\binom{n}{n}x^n$$

and $$(x+1)^n = \binom{n}{0}x^n+\binom{n}{1}x^{n-1}+\binom{n}{2}x^{n-1}+........+\binom{n}{n}$$

Now Coefficients of $x^n$ in $\displaystyle (1+x)^{2n} = \binom{2n}{n} = \binom{n}{0}^2+\binom{n}{1}^2+\binom{n}{2}^2+.....+\binom{n}{n}^2$

Now Using $\bf{Cauchy\; Schwarz}$ Inequality

$$\left[\binom{n}{0}^2+\binom{n}{1}^2+\binom{n}{2}^2+.....+\binom{n}{n}^2\right]\cdot \left[1^2+1^2+....+1^2\right]\geq \left[\binom{n}{0}+\binom{n}{1}+\binom{n}{2}+.....+\binom{n}{n}\right]^2=2^{2n}=4^n>4n\forall n\geq 2,n\in \mathbb{N}$$

So $$\binom{2n}{n} = \binom{n}{0}^2+\binom{n}{1}^2+\binom{n}{2}^2+.....+\binom{n}{n}^2>\frac{4n}{n+1}\forall n\geq 2,n\in \mathbb{N}$$

My question is , Is my proof is right, If not then how can i solve it.

Also plz explain any shorter way, Thanks

Answer: 
We prove the stronger bound $\binom{2n}{n} \geq \frac{4^n}{2n+1}$, which is bigger than $\frac{4}{n+1}$ for all $n \geq 2$.

Note that the $2n+1$ binomial coefficients $\binom{2n}{0}$, $\binom{2n}{1}$, ..., $\binom{2n}{2n}$ have average
$$
\frac{\binom{2n}{0} + \binom{2n}{1} + \ldots + \binom{2n}{2n}}{2n+1} = \frac{2^{2n}}{2n+1} = \frac{4^n}{2n+1}.
$$
Since $\binom{2n}{n}$ is the largest of these binomial coefficients, it is surely "above average": $\binom{2n}{n} \geq \frac{4^n}{2n+1}$.
Imagine a population, $P$, consisting of $n$ men and $n$ women. You are tasked with putting together a committee of $n$ people from $P$. There are no restriction on the make-up of the committee (for instance, it can be unisex). How many different committees can you put together?

Align the men in one row, and the women in another one. You can create a committee consisting of all the men. You can create a committee consisting of all the women. You can create a committee consisting of all the women in the odd-numbered places and all the men in the even places. You can create a committee consisting of all the men in the odd-numbered places and all the women in even places.

We have just enumerated four different possible committees. This is not necessarily an exhaustive list of all the possible committees, but it is enough for us to be able to make the statement that
$$
\binom{2n}{n} \geq 4.
$$

But
$$
4 = \frac{4n}{n} > \frac{4n}{n+1}.
$$