A sixth-degree lower bound at fixed sum six

Proved

WorkbookSource.base_38993

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

lean-workbooksource-checked

Let $a,b,c\ge 0$ such that $a+b+c=6$ . Prove that: $100+5(a^2+b^2+c^2)-2(a^2b^2+b^2c^2+a^2c^2)-a^2b^2c^2\ge 0$

Source: InternLM Lean-Workbook, record lean_workbook_38993 (Apache-2.0). Complete source proposition preserved; proof developed independently.

Preamble

import Mathlib
open Real Nat

Formal statement

theorem WorkbookSource.base_38993 (a b c : ℝ) (ha : a ≥ 0) (hb : b ≥ 0) (hc : c ≥ 0) (hab : a + b + c = 6) : 100 + 5 * (a ^ 2 + b ^ 2 + c ^ 2) - 2 * (a ^ 2 * b ^ 2 + b ^ 2 * c ^ 2 + a ^ 2 * c ^ 2) - a ^ 2 * b ^ 2 * c ^ 2 ≥ 0  :=  by sorry

Source

https://huggingface.co/datasets/internlm/Lean-Workbook/blob/main/lean_workbook.json, record lean_workbook_38993; Apache-2.0

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