Proposition: $\pi (2n) \ge \dfrac{\log \binom{ 2n }{n} }{\log 2n}$
Since this is a follow up proposition to this one: How can we show that $\operatorname{ord}_{p}\left(\binom{2n}n\right) \le \frac{\log 2n}{\log p}$
We can use : $\operatorname{ord}_{p}\left(\binom{2n}n\right) \le \frac{\log 2n}{\log p}$
attempt :
$n \log 2 \le \sum_{p\le 2n}\left[\frac{\log 2n}{\log p}\right]\log p \le \sum_{p \le 2n} \log 2n = \pi(2n) \log 2n$
(don't see how to "finish", if this is correct at all... )