A pairwise squared-product bound at fixed sum one

Proved

WorkbookSource.plus_14282

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

lean-workbooksource-checked

Let $a$ , $b$ and $c$ be non-negative numbers such that $a+b+c=1$ . Prove that: $216(a^2b^2+a^2c^2+b^2c^2)\leq11((a-b)^2+(a-c)^2+(b-c)^2)+8$

Source: InternLM Lean-Workbook, record lean_workbook_plus_14282 (Apache-2.0). Complete source proposition preserved; proof developed independently.

Preamble

import Mathlib
open Real Nat

Formal statement

theorem WorkbookSource.plus_14282 (a b c : ℝ) (ha : 0 ≤ a) (hb : 0 ≤ b) (hc : 0 ≤ c) (hab : a + b + c = 1) : 216 * (a^2 * b^2 + a^2 * c^2 + b^2 * c^2) ≤ 11 * ((a - b)^2 + (a - c)^2 + (b - c)^2) + 8   :=  by sorry

Source

https://huggingface.co/datasets/internlm/Lean-Workbook/blob/main/lean_workbook.json, record lean_workbook_plus_14282; Apache-2.0

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