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

Question

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 $ab$)?

Answer 1

From Properties of Some Functions Connected To Prime Numbers - Mincu et al.

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 $(7,7)$.

The proof is similar to your reasoning, with some additional edge cases.