lean_workbook_plus_76015

Proved

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

After substituting $\frac{ac+bd}{2}=1$ we have: \n\n $(ad-bc)^2 + \frac{(ac+bd)^2}{2} \ge \frac{(ad+bc)(ac+bd)}{2}$ \n\n $2a^2d^2 + 2b^2c^2 + a^2c^2 + b^2d^2 \ge a^2cd + abd^2 + abc^2 + b^2cd + 2abcd$

Preamble

import Mathlib.Analysis.Complex.Basic

Formal statement

theorem lean_workbook_plus_76015  (a b c d : ℝ) :
  2 * a^2 * d^2 + 2 * b^2 * c^2 + a^2 * c^2 + b^2 * d^2 ≥ a^2 * c * d + a * b * d^2 + a * b * c^2 + b^2 * c * d + 2 * a * b * c * d   :=  by sorry

Source

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

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