I have to prove that for every $n \in \mathbb{N}_+$ $$\binom{2n}{n} \ge \frac{4^n}{2\sqrt{n}}$$
It works for n = 1. But for n+1 I don't know what to do after this:
$$\binom{2(n+1)}{n+1}=\frac{2n!(2n+1)(2n+2)}{n!(n+1)(n+1)!}\ge\frac{4^n}{2\sqrt{n}}\cdot\frac{(2n+1)(2n+2)}{(n+1)^2}=\frac{n\cdot 4^{n+1}+4^n}{2\sqrt{n}(n+1)}$$
We can prove a much tighter inequality with elementary means. We have: $$ \frac{1}{4^n}\binom{2n}{n}=\frac{2}{\pi}\int_{0}^{\pi/2}\left(\cos\theta\right)^{2n}\,d\theta \tag{A}$$ $$ \frac{2}{\pi}\int_{0}^{\pi/2}\left(\cos\theta\right)^{2n+1}\,d\theta = \frac{2\cdot 4^n}{(2n+1)\pi\binom{2n}{n}}\tag{B}$$ and the sequence $\{a_m\}_{m\geq 0}$ given by $a_m=\frac{2}{\pi}\int_{0}^{\pi/2}\left(\sin\theta\right)^m\,d\theta$ is midpoint-log-convex by the Cauchy-Schwarz inequality. $a_{2n-1}\cdot a_{2n+1}\geq a_{2n}^2$ implies that $$ \frac{2\cdot 4^n}{(2n+1)\pi\binom{2n}{n}}\cdot \frac{2\cdot 4^{n-1}}{(2n-1)\pi\binom{2n-2}{n-1}}\geq \left[\frac{1}{4^n}\binom{2n}{n}\right]^2 \tag{C}$$
$$ \frac{1}{n\left(n+\tfrac{1}{2}\right)\pi^2}\geq \left[\frac{1}{4^n}\binom{2n}{n}\right]^4 \tag{D}$$ and $a_{2n}\cdot a_{2n+2}\geq a_{2n+1}^2 $ implies that: $$ \frac{1}{4^n}\binom{2n}{n}\cdot\frac{1}{4^{n+1}}\binom{2n+2}{n+1}\geq\left[\frac{2\cdot 4^n}{(2n+1)\pi\binom{2n}{n}}\right]^2\tag{E}$$
$$ \left[\frac{1}{4^n}\binom{2n}{n}\right]^4\geq \frac{(n+1) }{\pi^2\left(n+\tfrac{1}{2}\right)^3}\tag{F}$$ hence by $(D)$ and $(F)$ we have: $$\boxed{ \frac{1}{\sqrt{\pi}}\sqrt[4]{\frac{n+1}{\left(n+\tfrac{1}{2}\right)^3}}\leq \frac{1}{4^n}\binom{2n}{n}\leq\frac{1}{\sqrt{\pi}}\sqrt[4]{\frac{1}{n\left(n+\tfrac{1}{2}\right)}}.}\tag{G}$$
By the definition of the binomial coefficients and by the induction hypothesis,
$$\binom{2n}{n}=4\left(1-\frac1{2n}\right)\binom{2n-2}{n-1}\ge4\left(1-\frac1{2n}\right)\frac{4^{n-1}}{2\sqrt{n-1}}=\left(1-\frac1{2n}\right)\sqrt{\frac{n}{n-1}}\frac{4^n}{2\sqrt n}.$$
Remains to show that
$$\left(1-\frac1{2n}\right)\sqrt{\frac{n}{n-1}}\ge1.$$
By squaring,
$$1-\frac1n+\frac1{4n^2}\ge\frac{n-1}n.$$