I am looking for a proof, or a reference, or a remark or a comment with the conjecture as follows:
Let $\pi(x)$ is Prime-counting function of $x$. For $x$, $y$ be two positive integer numbers for $x, y \ge 8$ then
$$\pi(x.y) \ge \pi(x).\pi(y)$$
Noted: I compute the conjecture is true with $8 \le x, y \le 10^9$
PS: This conjecture inspired from Second Hardy–Littlewood conjecture, for $x, y \ge 2$ then
$$\pi(x+y) \le \pi(x)+\pi(y)$$
Here is an answer to the exercise. Let us take $x$ and $y$ sufficiently large, and take constants $A$ and $B$ so that we can apply $At \lt \pi(t)\log t \lt Bt$. Then using these and rewriting the conjectured inequality, we will have it once we can show $$xy\log x \log y \gt (B^2/A)xy\log (xy)$$. Taking generous values of $B=1.3$ and $A=0.9$, and letting $s=\log x$ and $t=\log y$, we get the above if we have $st \gt 1.9 (s+t)$ which in turn is implied by $(s-2)(t-2) \gt 4$. So we have the result when both $x$ and $y$ are greater than $55 \gt e^4$. To bring this down, you can tighten the inequalities above, or examine the roughly 1200 cases when $55 \geq x \geq y \geq 8$. Using equivalence classes induced by $\pi()$ can reduce the case checking to less than 80 cases.
Gerhard "Done Waiting At The Hospital" Paseman, 2018.06.26.