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Claim. Let $f(z)=\prod_{\nu=1}^n(z-z_\nu)$ with every $\lvert z_\nu\rvert\le1$. Theorem 4 of Pommerenke, On metric properties of complex polynomials, Michigan Math. J. 8 (1961), no. 2, 97–115 (printed p. 101), states that the set $E=\{z:\lvert f(z)\rvert\le1\}$ contains a disk of radius $(2e)^{-1}n^{-2}$; the proof centers the disk at a zero of $f$ and places it in the component of $E$ through $0$. Every point of the open disk has $\lvert f\rvert<1$, so in the notation of Problem 1039

$$\rho(f)\ge\frac{1}{2en^2}$$

for every such $f$, and the minimal inradius $\rho_n=\inf_{\deg f=n}\rho(f)$ is at least $1/(2en^2)$. The paper introduces the theorem by recalling the question of Erdős, Herzog and Piranian whether $\rho\ge\mathrm{const}\cdot n^{-1}$ when the zeros lie in the closed unit disk, their Problem 3, and says that it proves only a weaker estimate. The proof combines the diameter bound of the paper's Theorem 3 for the component through $0$, hence a capacity above $1/4$, with the author's derivative bound $\lvert f'\rvert<2en^2$ on that component, and integrates from a zero to the nearest boundary point. The theorem is recorded on the result page Theorem 4 of the card Pommerenke 1961.

Covers. The lower bound $\rho_n\ge1/(2en^2)$. It settles neither the order of $\rho_n$ nor the second question, whether $\rho_n\gg1/n$; the bound was improved to order $1/(n\sqrt{\log n})$ on Krishnapur, Lundberg and Ramachandran 2025, and the order $1/n$ is the claim on Price 2026.

Depends on. No page of this wiki: the proof is self-contained in the paper, with the external inputs the result page names.

Acceptance. Refereed: the paper appeared in the Michigan Mathematical Journal in 1961 (volume 8, issue 2; the issue carries no month, so the page is dated to the first day of the publication year). The site's commentary credits the bound to this paper, cited as [Po61], but labels the problem OPEN, so the remark is not acceptance and no reviewed is listed. Nothing here rests on this project's own review.