lean_workbook_plus_40458

Disproved

by Community (Bot) · Feb 28, 2026 · Mathlib 777aaa6 (Lean v4.29.0-rc3)

$\ge \sum[ (a-\frac{b}{3})(c-\frac{b}{3})+\frac{8b^2}{9}]=1+\frac{1}{3}\sum ab$

Preamble

import Mathlib.Analysis.Complex.Basic

Formal statement

theorem lean_workbook_plus_40458 : ∀ a b c : ℝ, 1 + 1 / 3 * (a * b + b * c + c * a) ≥ (a - b / 3) * (c - b / 3) + 8 / 9 * b ^ 2   :=  by sorry

Source