Let $\pi(x) = |\{ p \le x : p \in P\}|$ denote the prime counting function $\pi:\mathbb R \rightarrow \mathbb N$ and $P$ the set of primes.
The equality $$x \le \left\lfloor \prod_{p_i\le x} 1+\frac {\ln(x)} {\ln(p_i)} \right\rfloor \le \left(1+\frac {\ln(x)} {\ln(2)}\right)^{\pi(x)}$$
is stated to be true.
It should follow immediately that for all $x \ge 8:$ $$\pi(x)\ge \frac {\ln(x)} {2\ln\ln(x)}$$
However, can someone explain why this last inequality follows from the above inequality ?
There is no thery number involved in the proof:
You have $$\left( 1+\frac{\ln x}{\ln 2} \right)^{\pi(x)}\geq x$$ Then, since $\ln$ is increasing, $$\begin{align} \pi(x)\ln(1+\log_2 x)&\geq\ln x\\ \pi(x)&\geq\frac{\ln x}{\ln(1+\log_2 x)} \end{align}$$ so you are done if you show that $\ln(1+\log_2 x)\leq2\ln\ln x$, that is, $1+\log_2 x\leq(\ln x)^2$.
Define $f(t)=\left(t-\frac1t\right)\ln 2$. Differentiating you see easily that $f$ is increasing and with a calculator you can check that $f(\ln 8)>1$. Thus, for $x\geq 8$ you have $$\left(\ln x-\frac1{\ln x}\right)\ln 2=\left(\frac{(\ln x)^2-1}{\ln x}\right)\ln 2\geq 1$$ hence, $$(\ln x)^2-1\geq \log_2 x$$