A quadratic-sum product is at most two at fixed total two

Proved

WorkbookSource.base_23467

by wamlart · Sep 13, 2026 · Mathlib 0df444a (Lean v4.33.1)

lean-workbooksource-checked

Its easy to see that: $(ab+bc+ac-1)^2\geq0$ is true for all $a$ , $b$ and $c$ reals numbers.

$\Leftrightarrow1\geq(2-(ab+bc+ac))(ab+bc+ac)$

$\Leftrightarrow2\geq(4-2(ab+bc+ac))(ab+bc+ac)$

But when $a+b+c=2$ , $4-2(ab+bc+ac)=a^2+b^2+c^2$

$\Leftrightarrow2\geq(a^2+b^2+c^2)(ab+bc+ac)$

Source: InternLM Lean-Workbook, record lean_workbook_23467 (Apache-2.0). Complete source proposition preserved; proof developed independently.

Preamble

import Mathlib
open Real Nat

Formal statement

theorem WorkbookSource.base_23467  (a b c: ℝ)
  (h₀ : a + b + c = 2) :
  2 ≥ (a^2 + b^2 + c^2) * (a * b + b * c + c * a)  :=  by sorry

Source