For any natural number $x\ge 20$, we have that $$\pi(x)>\frac{\ln x}{2\ln\ln x}$$
, where $\pi(x)=$ the number of the set $\{p\in{\mathbb N}:p~ is~ a~prime~with~p\le x\}$ for each fixed $x\in{\mathbb R}$, which is so called prime-counting function .
My attempt :
Fixed positive integer $x\ge 20$ and let $p_{i}$ runs over all the primes less than or equal to $x$, that is, let $p_{i}\in\pi(x)$ for all $i$.
As $x\in{\mathbb N}$, then we have that $$x=p_{i_{1}}^{\alpha_{i_{1}}}p_{i_{2}}^{\alpha_{i_{2}}}\cdots p_{i_{r}}^{\alpha_{i_{r}}}=\prod_{k=1}^{r}p_{i_{k}}^{\alpha_{i_{k}}}$$, where each $p_{i_{k}}\in\pi(x)$ and each $\alpha_{i_{k}}\in{\mathbb N}.$
Since each $p_{i_{k}}^{\alpha_{i_{k}}}\le x$, one has for $x\ge 20$ that $$\alpha_{i_{k}}\le \frac{\ln x}{\ln p{_{i_{k}}}}\le 1+\frac{\ln x}{\ln p{_{i_{k}}}}\le 1+\frac{\ln x}{\ln 2}<\left(\ln x\right)^{2}$$ For the last inequality : note that \begin{align*} \left(\ln x\right)^{2}-\frac{\ln x}{\ln 2}&=\ln x\left(\ln x-\frac{1}{\ln 2}\right)\\ &=\ln x\left(\frac{\ln 2\ln x-1}{\ln 2}\right) \\ &\ge\ln 20\left(\frac{\ln 2\ln 20-1}{\ln 2}\right)\\ &>1\\ \end{align*} , where I use calculator to compute this value $\displaystyle\ln 20\left(\frac{\ln 2\ln 20-1}{\ln 2}\right)$ and the first inequality holds by $ \ln x$ is an increasing function.
Therefore, one has that $$x=\prod_{k=1}^{r}p_{i_{k}}^{\alpha_{i_{k}}}<\prod_{k=1}^{r}p_{i_{k}}^{\big(\ln x\big)^{2}}$$
Now, I get the stuck. I have no ideal how to proceed, or is there another way that can prove this theorem ?
Any comment or advice I will be grateful.