A weighted cyclic cubic ratio bounds the quadratic sum

Proved

WorkbookSource.base_47396

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

lean-workbooksource-checked

Prove that for positive reals a,b,c the following inequality holds good - $a^2(\frac {a + 2c}{3b}) + b^2(\frac {b + 2a}{3c}) + c^2(\frac {c + 2b}{3a}) \ge a^2 + b^2 + c^2$ - (own)

Source: InternLM Lean-Workbook, record lean_workbook_47396 (Apache-2.0). Complete source proposition preserved; proof developed independently.

Preamble

import Mathlib
open Real Nat

Formal statement

theorem WorkbookSource.base_47396 (a b c : ℝ) (ha : 0 < a) (hb : 0 < b) (hc : 0 < c) : a^2 * (a + 2 * c) / (3 * b) + b^2 * (b + 2 * a) / (3 * c) + c^2 * (c + 2 * b) / (3 * a) ≥ a^2 + b^2 + c^2  :=  by sorry

Source

https://huggingface.co/datasets/internlm/Lean-Workbook/blob/main/lean_workbook.json, record lean_workbook_47396; Apache-2.0

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