A squared quadratic ratio inequality with a normalized correction

Proved

WorkbookSource.plus_79615

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

lean-workbooksource-checked

For a,b,c positive reals $\quad \frac{(a^2+b^2+c^2)^2}{2(a^2b^2+b^2c^2+c^2a^2)+abc(a+b+c)}+1\geq \frac{6(a^2+b^2+c^2)}{(a+b+c)^2}$

P.S. Look at http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=384682 for the birth of it

Source: InternLM Lean-Workbook, record lean_workbook_plus_79615 (Apache-2.0). Complete source proposition preserved; proof developed independently.

Preamble

import Mathlib
open Real Nat

Formal statement

theorem WorkbookSource.plus_79615 (a b c : ℝ) (ha : 0 < a) (hb : 0 < b) (hc : 0 < c) : (a^2 + b^2 + c^2)^2 / (2 * (a^2 * b^2 + b^2 * c^2 + c^2 * a^2) + a * b * c * (a + b + c)) + 1 ≥ 6 * (a^2 + b^2 + c^2) / (a + b + c)^2   :=  by sorry

Source

https://huggingface.co/datasets/internlm/Lean-Workbook/blob/main/lean_workbook.json, record lean_workbook_plus_79615; Apache-2.0

View graph