A shifted mixed cyclic ratio sum is at least three

Proved

WorkbookSource.plus_3059

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

lean-workbooksource-checked

Let be $a,b,c > 0$ such that $a + b + c = 3$ . Prove that :

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

Source: InternLM Lean-Workbook, record lean_workbook_plus_3059 (Apache-2.0). Complete source proposition preserved; proof developed independently.

Preamble

import Mathlib
open Real Nat

Formal statement

theorem WorkbookSource.plus_3059 (a b c : ℝ) (ha : 0 < a) (hb : 0 < b) (hc : 0 < c) (hab : a + b + c = 3) : (a^2 * b + 2 * a^2 + 1) / (3 * b + 1) + (b^2 * c + 2 * b^2 + 1) / (3 * c + 1) + (c^2 * a + 2 * c^2 + 1) / (3 * a + 1) ≥ 3   :=  by sorry

Source

https://huggingface.co/datasets/internlm/Lean-Workbook/blob/main/lean_workbook.json, record lean_workbook_plus_3059; Apache-2.0

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