lean_workbook_plus_5318

Proved

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

Prove that $a^2-ab+b^2\ge \frac {a^2+ab+b^2} {3}$ which is equivalent to $(a-b)^2\ge 0$ assuming $a,b>0$ .

Preamble

import Mathlib.Analysis.Complex.Basic

Formal statement

theorem lean_workbook_plus_5318 (a b : ℝ) (ha : a > 0) (hb : b > 0) : a^2 - a * b + b^2 ≥ (a^2 + a * b + b^2) / 3   :=  by sorry

Source