A pair of cubic reciprocal ratios bounds pairwise ratios

Proved

WorkbookSource.base_8366

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

lean-workbooksource-checked

Let $a,b,c$ be positive real numbers. Prove that

\frac{a^3+b^3+c^3} {2abc}+\frac{2abc}{a^3+b^3+c^3} \ge \frac{a+b}{b+c}+\frac{b+c}{a+b}+\frac{1}{6}

Source: InternLM Lean-Workbook, record lean_workbook_8366 (Apache-2.0). Complete source proposition preserved; proof developed independently.

Preamble

import Mathlib
open Real Nat

Formal statement

theorem WorkbookSource.base_8366 (a b c : ℝ) (ha : 0 < a) (hb : 0 < b) (hc : 0 < c) : (a^3 + b^3 + c^3) / (2 * a * b * c) + (2 * a * b * c) / (a^3 + b^3 + c^3) ≥ (a + b) / (b + c) + (b + c) / (a + b) + 1 / 6  :=  by sorry

Source

https://huggingface.co/datasets/internlm/Lean-Workbook/blob/main/lean_workbook.json, record lean_workbook_8366; Apache-2.0

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