A weighted cubic ratio sum bounds five halves of the total

Proved

WorkbookSource.plus_64869

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

lean-workbooksource-checked

For $a, b, c>0$ prove that $\frac{3a^3+2b^3}{a^2+b^2}+\frac{3b^3+2c^3}{b^2+c^2}+\frac{3c^3+2a^3}{c^2+a^2}\ge\frac{5}{2}(a+b+c)$

Source: InternLM Lean-Workbook, record lean_workbook_plus_64869 (Apache-2.0). Complete source proposition preserved; proof developed independently.

Preamble

import Mathlib
open Real Nat

Formal statement

theorem WorkbookSource.plus_64869 (a b c : ℝ) (ha : 0 < a) (hb : 0 < b) (hc : 0 < c) : (3 * a ^ 3 + 2 * b ^ 3) / (a ^ 2 + b ^ 2) + (3 * b ^ 3 + 2 * c ^ 3) / (b ^ 2 + c ^ 2) + (3 * c ^ 3 + 2 * a ^ 3) / (c ^ 2 + a ^ 2) ≥ 5 / 2 * (a + b + c)   :=  by sorry

Source

https://huggingface.co/datasets/internlm/Lean-Workbook/blob/main/lean_workbook.json, record lean_workbook_plus_64869; Apache-2.0

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