\documentclass[11pt]{article} \usepackage[a4paper,margin=2.5cm]{geometry} \usepackage{amsmath,amssymb,amsthm} \usepackage{booktabs} \usepackage[hidelinks]{hyperref} \renewcommand{\labelitemi}{$\bullet$} \newtheorem{theorem}{Theorem} \newtheorem{lemma}{Lemma} \theoremstyle{remark}\newtheorem*{remark}{Remark} \title{The Schur--Siegel--Smyth trace problem:\\ the lower bound $1.8022246$} \author{David Niedbala Giraudin\\ \small Independent researcher, France --- ORCID 0009-0009-1526-1178} \date{27 October 2026 (this version: 10 November 2026)} \begin{document} \maketitle \begin{abstract} We prove $\lambda_{\rm SSS}\ge 9011123/5000000=1.8022246$ for the smallest limiting trace-to-degree ratio of totally positive algebraic integers. The previous best bound is $1.8022225$ (Tomar). The certificate consists of fourteen integer polynomials with rational multipliers and a continuous piecewise-linear density with $8\,678$ rational hat weights; its dual inequality is verified on $(0,\infty)$ in ball arithmetic. For the same family of polynomials the value of the underlying optimisation problem is $1.8022247$ to within $10^{-9}$ (numerical), so the bound is within $1.1\cdot10^{-7}$ of what this family can give. \end{abstract} \section{Statement} For a totally positive algebraic integer $\alpha$ of degree $d$ write $\overline{\rm tr}(\alpha)=\mathrm{tr}(\alpha)/d$, and let $\lambda_{\rm SSS}=\liminf\overline{\rm tr}(\alpha)$. \begin{theorem}[PROVED, computer-assisted]\label{thm} $\lambda_{\rm SSS}\ \ge\ \dfrac{9011123}{5000000}=1.8022246.$ \end{theorem} \noindent Previous lower bounds: $1.80203$ by Orloski, Talebizadeh Sardari and Smith~\cite{OSS}; $1.80220$ in~\cite{NG}; $1.8022225$ by Tomar~\cite{Tomar}. Several upper bounds obtained from Smith's theorem~\cite{Smith} have been announced since October 2026, for instance $1.8070809$~\cite{Tomar} and $1.80560940$~\cite{Pinho}. \section{The criterion} \begin{lemma}[\cite{OSS}; Proposition~2 of \cite{NG}; Lemma~2.1 of \cite{Tomar}]\label{lem} Let $Q_1,\dots,Q_k\in\mathbb Z[x]$ be nonzero, $\lambda_{Q}\ge0$, and let $g\ge0$ be a bounded density with compact support and mass $s>0$. Put $U_g(x)=\int\log|x-y|\,g(y)\,dy$ and $I(g)=\int U_g\,g$. If \[ F(x)=x-\lambda-\sum_Q\lambda_Q\log|Q(x)|-2U_g(x)+\frac{I(g)}{s}\ \ge\ 0 \] for every $x>0$ that is not a root of one of the $Q$, then $\lambda_{\rm SSS}\ge\lambda$. \end{lemma} \section{The certificate} The polynomials and multipliers are listed in Table~\ref{tab}; the polynomial $x^3-6x^2+9x-1$ receives multiplier $0$, so fourteen polynomials are used. The set is the one of~\cite{Tomar}. The density is \[ g=\sum_{k=1}^{8678}\rho_k h_k, \] where $h_k$ is the hat function of mass one on $[e_{k-1},e_{k+1}]$ with peak at $e_k$, for nodes $e_0<\dots