Prove2Me

lean_workbook_plus_47129

Proved

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

From the condition we have (a+b+c)(a2+b2+c2−ab−bc−ca)=1(a+b+c)(a^2+b^2+c^2-ab-bc-ca)=1 . Therefore a2+b2+c2=1a+b+c+ab+bc+caa^2+b^2+c^2= \frac{1}{a+b+c}+ab+bc+ca .

Preamble
import Mathlib.Analysis.Complex.Basic
Formal statement
theorem lean_workbook_plus_47129  (a b c : ℂ)
  (h₀ : (a + b + c) * (a^2 + b^2 + c^2 - a * b - b * c - c * a) = 1) :
  a^2 + b^2 + c^2 = 1 / (a + b + c) + a * b + b * c + c * a   :=  by sorry
Source
https://huggingface.co/datasets/internlm/Lean-Workbook

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