A cubed total bounds a cubic sum with a triple-product correction

Proved

WorkbookSource.base_40599

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

lean-workbooksource-checked

Let $a,b,c> 0$ . Prove:

(a+b+c)^3(a^3+b^3+c^3+abc)\ge 36abc(a^3+b^3+c^3)

To flvbin: solution by hand

$LHS - RHS$

equals to:

$\sum a^4(a-b)(a-c) +3(a-b)^2(b-c)^2(c-a)^2+ \sum [c^3+(ab^2+a^2b+c^3-3abc)] (a-b)^2 \geq 0$

Source: InternLM Lean-Workbook, record lean_workbook_40599 (Apache-2.0). Complete source proposition preserved; proof developed independently.

Preamble

import Mathlib
open Real Nat

Formal statement

theorem WorkbookSource.base_40599 (a b c : ℝ) (ha : 0 < a) (hb : 0 < b) (hc : 0 < c) : (a + b + c) ^ 3 * (a ^ 3 + b ^ 3 + c ^ 3 + a * b * c) ≥ 36 * a * b * c * (a ^ 3 + b ^ 3 + c ^ 3)  :=  by sorry

Source

https://huggingface.co/datasets/internlm/Lean-Workbook/blob/main/lean_workbook.json, record lean_workbook_40599; Apache-2.0

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