Given that $m,s \in \mathbb{N}$, if $m \ge 8 s \log(m^2 s)$, how much greater $m$ is relatively to $s$ ?
It seems to me $m>>s$, but I would like some idea of the magniture. I'm not quite sure how to attack this problem.
This is obviously equivalent to $m \ge 16 s \log(m) + 8 s \log(s)$.
Maybe an idea is to see how much $m$ has to grow when $s$ increases by 1...
Setting $m::=16ts$, the inequation turns to
$$16ts\ge8s\log(256t^2s^3),$$ or, dividing by $16s$ $$t\ge4\log2+\log t+\frac32\log s$$ and taking the exponential, $$\frac{e^t}t\ge16s^{3/2},$$ or $$(-t)e^{-t}\ge-\frac1{16s^{3/2}}.$$
Then, using Lambert's W function (secondary branch, decreasing),
$$-t\le W\left(-\frac1{16s^{3/2}}\right),$$ $$m\ge-16W\left(-\frac1{16s^{3/2}}\right)s.$$