lean_workbook_plus_36660

Proved

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

Let $a, b, c>0$ and $\frac{1}{a^2(b+c)} + \frac{1}{b^2(a+c)} + \frac{1}{c^2(a+b)}=1$ . Prove that

abc \ge \frac{3}{2}

Preamble

import Mathlib.Analysis.Complex.Basic

Formal statement

theorem lean_workbook_plus_36660 (a b c : ℝ) (ha : 0 < a) (hb : 0 < b) (hc : 0 < c) (habc : a * b * c = 1) (h : (1 / (a^2 * (b + c)) + 1 / (b^2 * (a + c)) + 1 / (c^2 * (a + b))) = 1) : a * b * c ≥ 3 / 2   :=  by sorry

Source

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

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