lean_workbook_plus_559

Proved

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

By squaring and adding side by side them: $|a|\ge |b+c|,|b|\ge |c+a|$ and $|c|\ge |a+b|$ we get $(a+b+c)^2\le 0 \Longrightarrow \ \ \ a+b+c=0$

Preamble

import Mathlib.Analysis.Complex.Basic

Formal statement

theorem lean_workbook_plus_559 (a b c : ℝ) (h : |a| ≥ |b + c| ∧ |b| ≥ |c + a| ∧ |c| ≥ |a + b|) :
  a + b + c = 0   :=  by sorry

Source