Combinatorial proof of $\frac{2^{2n}}{n+1}$ $\leq$ ${2n}\choose{n}$ I have proved this inequality with the help of the $RMS-AM$ inequality which states that $\sqrt\frac{{x_0}^2+{x_1}^2+...+{x_n}^2}{n+1}\geq\frac{x_0+x_1+...+x_n}{n+1}\implies\frac{{x_0}^2+{x_1}^2+...+{x_n}^2}{n+1}\geq\frac{(x_0+x_1+...+x_n)^2}{(n+1)^2}$. Now, if we replace $x_0$ by ${n}\choose{0} $, $x_1$ by $n \choose 1$ and so on, we get ${2n}\choose{n}$ $\geq$ $\frac{2^{2n}}{n+1}$
I was wondering if a combinatorial proof can be found for this inequality. 
 A: Embarrassingly, my original answer (which got accepted and upvoted, and which I have deleted now) was totally wrong. Here is a correct, one-line proof - but not a completely combinatorial one: 
$$ \binom{2n}n = \sum_{k=0}^n {\binom nk}^2 \ge \frac1{n+1} \Bigg( \sum_{k=0}^n \binom nk \Bigg)^2 = \frac{2^{2n}}{n+1}. $$
Here the first equality is a well-known identity expressing in two different ways the number of binary strings of length $2n$ with exactly $n$ components equal to $1$, and the middle inequality is Cauchy-Schwarz. 

Incidentally, the particular case of the Cauchy-Schwarz inequality used above can be established using a combinatorial-style argument. The starting point is the fact that the "scalar product" of two sets with non-negative elements is maximized when the sets are similarly ordered. If we accept this fact as combinatorial, we proceed to conclude that for any permutation $\pi\in S_{n+1}$,
  $$ \sum_{k=0}^n {\binom nk}^2 \ge \sum_{k=0}^n \binom nk \binom n{\pi(k)}. $$
Averaging over all $(n+1)!$ permutations, and observing that for each $j\in[0,n]$, the product $\binom nk\binom nj$ appears in the RHS for exactly $n!$ permutations, gives the result.
