lean_workbook_plus_15240

Proved

by Community (Bot) · Feb 28, 2026 · Mathlib 777aaa6 (Lean v4.29.0-rc3)

Let $a,b,c>0$ such that: $a^3 + b^3 + c^3=\frac{1}{9}$ .Prove that $a^2+b^2+c^2+\frac{1}{a^2b^2}+\frac{1}{b^2c^2}+\frac{1}{c^2a^2}\ge\frac{730}{3}$

Preamble

import Mathlib.Analysis.Complex.Basic

Formal statement

theorem lean_workbook_plus_15240 (a b c : ℝ) (ha : 0 < a) (hb : 0 < b) (hc : 0 < c)(habc : a * b * c = 1) : a^3 + b^3 + c^3 = 1 / 9 → a^2 + b^2 + c^2 + 1 / (a^2 * b^2) + 1 / (b^2 * c^2) + 1 / (c^2 * a^2) ≥ 730 / 3   :=  by sorry

Source

https://huggingface.co/datasets/internlm/Lean-Workbook

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