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A product of cyclic cubic sums bounds a cubed sum

Proved
WorkbookSource.base_34880

by wamlart · Sep 12, 2026 · Mathlib 0df444a (Lean v4.33.1)

lean-workbooksource-checked

Given x,y,z≥0 x, y, z \geq 0x,y,z≥0, prove the inequality: 3(x2y+y2z+z2x)(xy2+yz2+zx2)≥xyz(x+y+z)3 3(x^2y + y^2z + z^2x)(xy^2 + yz^2 + zx^2) \ge xyz(x + y + z)^33(x2y+y2z+z2x)(xy2+yz2+zx2)≥xyz(x+y+z)3

Source: InternLM Lean-Workbook, record lean_workbook_34880 (Apache-2.0). Complete source proposition preserved; proof developed independently.

Preamble
import Mathlib
open Real Nat
Formal statement
theorem WorkbookSource.base_34880 (x y z : ℝ) (hx : x ≥ 0) (hy : y ≥ 0) (hz : z ≥ 0) : 3 * (x^2 * y + y^2 * z + z^2 * x) * (x * y^2 + y * z^2 + z * x^2) ≥ x * y * z * (x + y + z)^3  :=  by sorry
Source
https://huggingface.co/datasets/internlm/Lean-Workbook/blob/main/lean_workbook.json, record lean_workbook_34880; Apache-2.0

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