# Proving $\pi(ab) \geq \pi(a)\pi(b)$?

It occurs to me that a stronger version of my related post the other day would be:

$$\pi(ab) \geq \pi(a) \cdot \pi(b)\ ,$$

which I believe holds for all naturals $$a,b$$ except for the two pairs $$(5,7)$$ and $$(7,7)$$.

My first thought on a proof was to point out that

$$\pi(ab)\sim\frac{ab}{\log(ab)}>\frac{ab}{\log(a)\log(b)}\sim \pi(a)\pi(b)\ ,$$

given $$a>e$$ and $$b>a^{1-\log{a}}$$ to rule out the small log problems, though I'm not sure that's necessary.

My question is whether that's enough to establish that the conjecture is always true for sufficiently large $$ab$$, or if it merely implies that it will tend towards being true in the limit. If this isn't enough and someone has a different way to go at it, I'd be interested to hear it.

And if this approach is valid, is it safe to change the conjecture to the stricter $$\pi(ab)>\pi(a)\pi(b)$$ (for large enough $$ab$$)?

Theorem 3.4: For $$x,y\geq \sqrt{53}$$, $$\pi(x)\pi(y)\leq \pi(xy)$$
Remark 6.6 The relation $$π(xy) ≥ π(x)π(y)$$ holds for all positive integers $$x, y$$ with the following three exceptions: x = 5, y = 7; x = 7, y = 5 and x = y = 7.
So note the additional exception of $$(7,7)$$.