Show $$\pi(x) \ll \frac{x \log \log x}{\log x}.$$
I was able to show a simpler expression, $$\pi (x) \ll \frac{x}{\log \log x}$$ in hopes of using this to prove the original statement, but I still do not have luck. I will attach the proof of $\pi (x) \ll \frac{x}{\log \log x}$ (to avoid making it too long, I omitted some steps, which I can add if requested).
If $r$ is so that $p_1, \cdots , p_r$ are all primes smaller than $\sqrt{x}$ then we obtain \begin{align*} \pi (x) & \leq 2^{r+1} + x\prod_{i=1}^r \left(1 - \frac{1}{p_i} \right) \\ & \leq 2^{r+1} + \frac{x}{\log p_r} \\ & \leq 2^{\log x + 2} + \frac{x}{\log \log x} \\ & \leq 4 \cdot 2^{\log x} + \frac{x}{\log \log x} \\ & = O(2 ^{\log x}) + \frac{x}{\log \log x} \\ & \leq o\left( \frac{x}{\log \log x} \right) + \frac{x}{\log \log x} \\ & = O\left( \frac{x}{\log \log x} \right). \end{align*} Thus, $$\pi (x) \ll \frac{x}{\log \log x}.$$