How can the estimate $ \binom{n}{\frac {n}{2}} \ge \frac {2^n}{n+1} $be made? Let's define the average value of binomial coefficients by $$
A (n) = \frac {1}{n+1} \sum_{i=0}^n \binom{n}{i} 
$$
Clearly, $$A= \frac {2^n}{n+1} \qquad .$$ 
Now, this gives us the trivial estimate for $$ \binom{n}{\frac {n}{2}} $$ as
$$ \binom{n}{\frac {n}{2}} \ge \frac {2^n}{n+1}
$$
I know that the largest possible value of $\binom{n}{k}$ is at $k= \frac {n}{2}$ 
The estimate holds just because this largest value is greater than the average value of binomial coefficients?
Is it enough to say that this estimation holds since the maximum number in a set is always at least the average value?

Just induction!
Firstly, we'll replace $n$ at $2n$.
Thus, we need to prove that $\binom{2n}{n}\geq\frac{2^{2n}}{2n+1}.$
For $n=1$ it's obvious.
Let $$\binom{2n}{n}\geq\frac{2^{2n}}{2n+1}.$$
Hence,$$\binom{2n+2}{n+1}=\binom{2n}{n}\cdot\frac{(2n+1)(2n+2)}{(n+1)^2}\geq\frac{2^{2n}}{2n+1}\cdot\frac{2(2n+1)}{n+1}=$$
$$=\frac{2^{2n+2}}{2n+3}\cdot\frac{2n+3}{2(n+1)}>\frac{2^{2n+2}}{2n+3}$$
and we are done!
