lean_workbook_plus_19961

Proved

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

Given the inequality $\frac {a^2b^2+a^2(a-b)^2+b^2(a-b)^2}{a^2b^2(a-b)^2} \ge \frac {4}{3+2a+2b+ab}$ for positive real numbers a and b with $a \neq b$ , prove that it is true.

Preamble

import Mathlib.Analysis.Complex.Basic

Formal statement

theorem lean_workbook_plus_19961 (a b : ℝ) (ha : 0 < a) (hb : 0 < b) (hab : a ≠ b) : (a^2 * b^2 + a^2 * (a - b)^2 + b^2 * (a - b)^2) / (a^2 * b^2 * (a - b)^2) ≥ 4 / (3 + 2 * a + 2 * b + a * b)   :=  by sorry

Source

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

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