lean_workbook_plus_60671

Proved

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

prove that $(a^2+b^2+c^2)\cdot\frac{(p-1)^2}{2}\ge(ab+bc+ca)\cdot\frac{(p-1)^2}{2}$ for $a,b,c,p\ge 0$

Preamble

import Mathlib.Analysis.Complex.Basic

Formal statement

theorem lean_workbook_plus_60671 (a b c p : ℝ) (ha : a ≥ 0) (hb : b ≥ 0) (hc : c ≥ 0) (hp : p ≥ 0) : (a^2 + b^2 + c^2) * (p - 1)^2 / 2 ≥ (a * b + b * c + c * a) * (p - 1)^2 / 2   :=  by sorry

Source

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

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