A cyclic quadratic ratio bound with a difference correction

Proved

WorkbookSource.base_7718

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

lean-workbooksource-checked

Let $a, b, c>0$ and $a+b+c=3$ .

\frac{a^2}{b(c+a)} +\frac{b^2}{c(a+b)}+\frac{c^2}{a(b+c)}\geq \frac{3}{2}+\frac{(a-b) (b-c) (c-a)} {abc}

Source: InternLM Lean-Workbook, record lean_workbook_7718 (Apache-2.0). Complete source proposition preserved; proof developed independently.

Preamble

import Mathlib
open Real Nat

Formal statement

theorem WorkbookSource.base_7718 (a b c : ℝ) (ha : 0 < a) (hb : 0 < b) (hc : 0 < c) (hab : a + b + c = 3) : a^2 / (b * (c + a)) + b^2 / (c * (a + b)) + c^2 / (a * (b + c)) ≥ 3 / 2 + (a - b) * (b - c) * (c - a) / (a * b * c)  :=  by sorry

Source

https://huggingface.co/datasets/internlm/Lean-Workbook/blob/main/lean_workbook.json, record lean_workbook_7718; Apache-2.0

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