Prove that $\binom{2n}{n}>\frac{4n}{n+1}\forall \; n\geq 2, n\in \mathbb{N}$ 
Prove that $$\binom{2n}{n}>\frac{4n}{n+1}\forall \; n\geq 2, n\in \mathbb{N}$$

$\bf{My\; Try::}$ 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
 A: 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": $\binom{2n}{n} \geq \frac{4^n}{2n+1}$.
A: The proof is correct. I particularly enjoyed how you proved the identity involving the squares of the binomial coefficients.
A: 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
$$
4 = \frac{4n}{n} > \frac{4n}{n+1}.
$$
A: It can probably be made shorter using induction: $$\binom{2(k+1)}{k+1)}=\binom{2k}k\cdot \frac{(2k+1)(2k+2)}{k+1}\\
>\frac{4k(2k+1)(2k+2)}{(k+1)^2}$$It isn't very hard to show that the final fraction is larger than $\frac{4k+4}{k+2}$.
A: Some other approaches look like sledgehammers to me, as you essentially need to show that the central binomial coefficient exceeds $4$, which is true for $2n=4$.
Is it really necessary to prove that the central coefficients are increasing ?

If yes,
$$\binom{2n+2}{n+1}=\frac{(2n+2)(2n+1)}{(n+1)(n+1)}\binom{2n}n$$
or simply invoke Pascal's identity that shows that the central element at least doubles on every second row.

Note that the first product shows that the value is asymptotically mutiplied by $4$ on every other row.
A: $$\binom{2n}n=\frac{n+1}{1}\frac{n+2}{2}\cdots\frac{2n-1}{n-1}\frac{2n}{n}\ge2^n>\frac{4n}{n+1}.$$
