lean_workbook_plus_13853

Proved

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

For $x\ge y\ge z$ or $x\ge z\ge y$ and $x,y,z>0$ , prove that $x^3+y^3+z^3+2(x^2y+y^2z+z^2x)\ge 3(x^2y+y^2z+z^2x)$ .

Preamble

import Mathlib.Analysis.Complex.Basic

Formal statement

theorem lean_workbook_plus_13853 (x y z : ℝ) (h : x ≥ y ∧ y ≥ z ∨ x ≥ z ∧ z ≥ y) (hx : 0 < x ∧ 0 < y ∧ 0 < z) : x^3 + y^3 + z^3 + 2 * (x^2 * y + y^2 * z + z^2 * x) ≥ 3 * (x^2 * y + y^2 * z + z^2 * x)   :=  by sorry

Source

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

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