Inequality of factorial, prove ${2n\choose n} \ge {n \choose {\left\lfloor \frac{n}{2} \right\rfloor}}$. my name is Rafał and I decided to create this thread because of my inability to find a solution. I have been fighting with this inequality for 1.5 week and I have a hope that you will give me any hint (no solution but clue). So I need to prove it to prove something bigger. I'm talking about inequality:

$${2n\choose n} \ge {n \choose {\left\lfloor \frac{n}{2} \right\rfloor}}$$

In my opinion it's obvious, but formally i need to prove it. 
1)So I wrote a simple program in C++ for n from 1 to 11 and everything was allright (bigger n implicited problem with memory). It was very clear to see that ${2n\choose n}$ was growing faster.  
2)I was trying to use definition of binomial coefficients and by using mathematical induction proved this inequality but there were such a big problems with ${\left\lfloor \frac{n}{2} \right\rfloor}$. (Yes, I used ${\frac{n}{2}-1 \le \left\lfloor \frac{n}{2} \right\rfloor} \le \frac{n}{2} \le \frac{n}{2}+1$ but it wasn't helpful)
3) I have tried to define a sequence like this: $a_n = \frac{{2n\choose n}}{{n \choose {\left\lfloor \frac{n}{2} \right\rfloor}}}$. My idea was to show that the sequence is increasing for every n ($\frac{a_{n+1}}{a_n} \ge 1$). When it'll be true this sequence must be bounded below by $a_1=1$ and it'll be the end of proof.
But every of my methods were incorect or maybe i had a problem with transformations. I have next idea to use Stirling approximation but I have completely no idea if it will be corect. Any Hint?
 A: Intuitively (and combinatorially), ${n \choose r}$ is the number of ways of forming strings of length $n$ over the alphabet $\{0,1\}$ where $r$ of the letters are $0$.
Every string counted by ${n \choose \lfloor{\frac{n}{2}}\rfloor }$ can be extended to a string counted by ${2n \choose n}$. Moreover, there are multiple ways of doing it. So the inequality follows.
A: $$ \binom{2n}{n} = \sum_{k=0}^n \binom{n}{k}^2 \ge \binom{n}{\lfloor\frac n2\rfloor}^2 \ge \binom{n}{\lfloor\frac n2\rfloor} $$
The first equality is proved in this question, for example.
The related inequalities
$$ \binom{2n}{n} < 2^n\binom{n}{n/2} < 2\binom{2n}{n} $$
are in this question.
Also, this question shows $(\frac nk)^k \le \binom nk$; thus
$$ \binom{2n}{n} \ge \left(\frac{2n}{n}\right)^n = 2^n \ge \binom n{\lfloor n/2\rfloor} $$
A: Setting $n=2m$, the inequation can be rewritten as$$(4m)!m!^2\ge(2m)!^3.$$
Let us expand for $m=3$:
$$12.11.10.9.8.7.6.5.4.3.2.1.(3.2.1)^2\ge(6.5.4.3.2.1)^3.$$
We first simplify by $(2m)!$:
$$12.11.10.9.8.7.(3.2.1)^2\ge(6.5.4.3.2.1)^2.$$
Next by $m!^2$:
$$12.11.10.9.8.7\ge(6.5.4)^2.$$
And by $(2m)!/m!$:
$$2.11.2.9.2.7\ge6.5.4.$$
Every remaining factor in the RHS has a corresponding larger factor in the LHS.
This generalizes to other $m$ and the same work needs to be repeated for odd $n=2m+1$.

Again for $m=3$:
$$14.13.12.11.10.9.8.7.6.5.4.3.2.1.(3.2.1)^2\ge(7.6.5.4.3.2.1)^3.$$
We first simplify by $(2m+1)!$:
$$14.13.12.11.10.9.8.(3.2.1)^2\ge(7.6.5.4.3.2.1)^2.$$
Next by $m!^2$:
$$14.13.12.11.10.9.8\ge(7.6.5.4)^2.$$
And by $(2m+1)!/m!$:
$$2.13.2.11.2.9.2\ge7.6.5.4.$$
