A normalized cyclic cubic inequality

Proved

WorkbookSource.base_15342

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

lean-workbooksource-checked

Prove that: $4 + 15abc + (a - b)(b - c)(c - a) \geq 5(a^2b + b^2c + c^2a) + 12(ab + bc + ca)$ given $a;b;c \geq 0$ and $a + b + c = 1$ .

Source: InternLM Lean-Workbook, record lean_workbook_15342 (Apache-2.0). Complete source proposition preserved; proof developed independently.

Preamble

import Mathlib
open Real Nat

Formal statement

theorem WorkbookSource.base_15342 (a b c : ℝ) (ha : 0 ≤ a) (hb : 0 ≤ b) (hc : 0 ≤ c) (hab : a + b + c = 1) : 4 + 15 * a * b * c + (a - b) * (b - c) * (c - a) ≥ 5 * (a^2 * b + b^2 * c + c^2 * a) + 12 * (a * b + b * c + c * a)  :=  by sorry

Source

https://huggingface.co/datasets/internlm/Lean-Workbook/blob/main/lean_workbook.json, record lean_workbook_15342; Apache-2.0

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