Prove using induction the inequality. 
$\forall$:n∈${N}$
$\binom{2n}{n}$ $\ge  \frac{4^n}{2n+1}$
I tried with no any success...

 A: $$n=1 :\binom{2}{1}=2 \geq \dfrac{4}{3}
\\n=k \to \binom{2k}{k}\geq \dfrac{4^k}{2k+1} \\
 n=k+1 \to \binom{2(k+1)}{k+1}\geq \dfrac{4^{k+1}}{2(k+1)+1}$$
$$\binom{2(k+1)}{k+1}=\dfrac{(2k+2)!}{(k+1)!(k+1)!}=\dfrac{(2k+2)(2k+1)(2k)!}{(k+1)^2k!}=\\ 
\dfrac{(2k+2)(2k+1)}{(k+1)^2} \binom{2k}{k}\geq \dfrac{4.4^k}{2k+3}=4.\dfrac{2k+1}{2k+3}\dfrac{4^k}{2k+1}\\$$if we prove the $\dfrac{(2k+2)(2k+1)}{(k+1)^2} \geq 4\dfrac{2k+1}{2k+3}$ ,we proved the original relation 
$$\dfrac{(2k+2)(2k+1)}{(k+1)^2} \geq 4\dfrac{2k+1}{2k+3} \to \\
\dfrac{(2k+2)}{(k+1)^2} \geq 4\dfrac{1}{2k+3}\\
\dfrac{2}{(k+1)} \geq 4\dfrac{1}{2k+3}\\\\2(2k+3) \geq 4(k+1) \\4k+6 \geq 4k+4\\6\geq 4 \checkmark$$
A: For $n=1$ we get $2\geq\frac{4}{3}$, which is true.
If $\binom{2n}{n}\geq\frac{4^n}{2n+1}$ we obtain:
$$\binom{2n+2}{n+1}=\binom{2n}{n}\cdot\frac{(2n+2)(2n+1)}{(n+1)^2}>\frac{4^n}{2n+1}\cdot\frac{2(2n+1)}{n+1}$$ and it remains to prove that
$$\frac{4^n}{2n+1}\cdot\frac{2(2n+1)}{n+1}>\frac{4^{n+1}}{2n+3}$$ or
$$2(2n+3)>4(n+1),$$
which is obvious.
