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: 
$$\binom{2n}n=\frac{n+1}{1}\frac{n+2}{2}\cdots\frac{2n-1}{n-1}\frac{2n}{n}\ge2^n>\frac{4n}{n+1}.$$
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 $\frac{4k+4}{k+2}$.