A cyclic cubic ratio bounds the quadratic sum at fixed sum three

Proved

WorkbookSource.base_36161

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

lean-workbooksource-checked

If the positive reals $a,b,c$ satisfy $a+b+c=3$ prove that

\sum_{cyc} \frac{a^3}{b^2}\geq a^2+b^2+c^2.

By computer,we have

LHS-RHS=\frac{1}{3}\sum_{cyc}{\frac{(ab+bc+3a+3b)(a-b)^2}{b^2}}\ge{0}

Source: InternLM Lean-Workbook, record lean_workbook_36161 (Apache-2.0). Complete source proposition preserved; proof developed independently.

Preamble

import Mathlib
open Real Nat

Formal statement

theorem WorkbookSource.base_36161 (a b c : ℝ) (ha : 0 < a) (hb : 0 < b) (hc : 0 < c) (hab : a + b + c = 3) : a^3 / b^2 + b^3 / c^2 + c^3 / a^2 ≥ a^2 + b^2 + c^2  :=  by sorry

Source

https://huggingface.co/datasets/internlm/Lean-Workbook/blob/main/lean_workbook.json, record lean_workbook_36161; Apache-2.0

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