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

 [/INST] 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.
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