A cyclic sixth-degree inequality in mixed powers

Proved

WorkbookSource.base_8279

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

lean-workbooksource-checked

Prove that $ab^5 + bc^5 + ca^5\ge abc(a^2b + b^2c + c^2a)$ , for $a,b,c > 0$ .

Source: InternLM Lean-Workbook, record lean_workbook_8279 (Apache-2.0). Complete source proposition preserved; proof developed independently.

Preamble

import Mathlib
open Real Nat

Formal statement

theorem WorkbookSource.base_8279 (a b c : ℝ) (ha : 0 < a) (hb : 0 < b) (hc : 0 < c) : a * b^5 + b * c^5 + c * a^5 ≥ a * b * c * (a^2 * b + b^2 * c + c^2 * a)  :=  by sorry

Source

https://huggingface.co/datasets/internlm/Lean-Workbook/blob/main/lean_workbook.json, record lean_workbook_8279; Apache-2.0

View graph