A cubic-over-quadratic lower bound at fixed sum six

Proved

WorkbookSource.base_11044

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

lean-workbooksource-checked

Prove that for positive numbers $a, b, c$ with $a+b+c=6$ , the following inequality holds: $\sum \frac{a^3}{a^2+b+c} \ge 3$

Source: InternLM Lean-Workbook, record lean_workbook_11044 (Apache-2.0). Complete source proposition preserved; proof developed independently.

Preamble

import Mathlib
open Real Nat

Formal statement

theorem WorkbookSource.base_11044 (a b c : ℝ) (ha : 0 < a) (hb : 0 < b) (hc : 0 < c) (hab : a + b + c = 6) : a^3 / (a^2 + b + c) + b^3 / (b^2 + c + a) + c^3 / (c^2 + a + b) ≥ 3  :=  by sorry

Source

https://huggingface.co/datasets/internlm/Lean-Workbook/blob/main/lean_workbook.json, record lean_workbook_11044; Apache-2.0

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