A fifth-power ratio bounds a corrected cubic sum

Proved

WorkbookSource.base_57179

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

lean-workbooksource-checked

Let $a,b,c>0.$ Prove that $\frac{a^5+b^5+c^5}{a^2+b^2+c^2}\ge \frac{1}{2}(a^3+b^3+c^3-abc)$

Source: InternLM Lean-Workbook, record lean_workbook_57179 (Apache-2.0). Complete source proposition preserved; proof developed independently.

Preamble

import Mathlib
open Real Nat

Formal statement

theorem WorkbookSource.base_57179 (a b c : ℝ) (ha : 0 < a) (hb : 0 < b) (hc : 0 < c) : (a^5 + b^5 + c^5) / (a^2 + b^2 + c^2) ≥ (1 / 2) * (a^3 + b^3 + c^3 - a * b * c)  :=  by sorry

Source

https://huggingface.co/datasets/internlm/Lean-Workbook/blob/main/lean_workbook.json, record lean_workbook_57179; Apache-2.0

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