A symmetric rational inequality involving squared pairwise products

Proved

WorkbookSource.base_6624

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

lean-workbooksource-checked

Prove that $\frac{abc}{a^2b^2+b^2c^2+c^2a^2}+\frac{2}{a+b+c}\ge \frac{a+b+c}{a^2+b^2+c^2}$ given $a,b,c>0$ . Is there a solution without full expanding?

Source: InternLM Lean-Workbook, record lean_workbook_6624 (Apache-2.0). Complete source proposition preserved; proof developed independently.

Preamble

import Mathlib
open Real Nat

Formal statement

theorem WorkbookSource.base_6624 (a b c : ℝ) (ha : 0 < a) (hb : 0 < b) (hc : 0 < c) : (a * b * c) / (a ^ 2 * b ^ 2 + b ^ 2 * c ^ 2 + c ^ 2 * a ^ 2) + 2 / (a + b + c) ≥ (a + b + c) / (a ^ 2 + b ^ 2 + c ^ 2)  :=  by sorry

Source

https://huggingface.co/datasets/internlm/Lean-Workbook/blob/main/lean_workbook.json, record lean_workbook_6624; Apache-2.0

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