A cubic symmetric bound at squared norm three

Proved

WorkbookSource.base_49447

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

lean-workbooksource-checked

For $a, b, c \ge 0$ and $a^2+b^2+c^2=3$ , prove that $a^3+b^3+c^3+6abc\leq 3(a+b+c)$ .

Source: InternLM Lean-Workbook, record lean_workbook_49447 (Apache-2.0). Complete source proposition preserved; proof developed independently.

Preamble

import Mathlib
open Real Nat

Formal statement

theorem WorkbookSource.base_49447 (a b c : ℝ) (ha : 0 ≤ a) (hb : 0 ≤ b) (hc : 0 ≤ c) (hab : a^2 + b^2 + c^2 = 3) : a^3 + b^3 + c^3 + 6 * a * b * c ≤ 3 * (a + b + c)  :=  by sorry

Source

https://huggingface.co/datasets/internlm/Lean-Workbook/blob/main/lean_workbook.json, record lean_workbook_49447; Apache-2.0

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