Prove2Me

A refined squared pairwise reciprocal lower bound

Proved
WorkbookSource.base_28399

by wamlart · Sep 13, 2026 · Mathlib 0df444a (Lean v4.33.1)

lean-workbooksource-checked

If x,y,z>0x,y,z>0 prove: 1(y+z)2+1(z+x)2+1(x+y)2≥(x+y+z)[9(x+y)(y+z)(z+x)−8xyz]4(x+y)2(y+z)2(z+x)2\frac{1}{(y+z)^2}+\frac{1}{(z+x)^2}+\frac{1}{(x+y)^2}\ge\frac{(x+y+z)[9(x+y)(y+z)(z+x)-8xyz]}{4(x+y)^2(y+z)^2(z+x)^2}

Source: InternLM Lean-Workbook, record lean_workbook_28399 (Apache-2.0). Complete source proposition preserved; proof developed independently.

Preamble
import Mathlib
open Real Nat
Formal statement
theorem WorkbookSource.base_28399 (x y z : ℝ) (hx : 0 < x) (hy : 0 < y) (hz : 0 < z) : (1 / (y + z) ^ 2 + 1 / (z + x) ^ 2 + 1 / (x + y) ^ 2) ≥ (x + y + z) * (9 * (x + y) * (y + z) * (z + x) - 8 * x * y * z) / (4 * (x + y) ^ 2 * (y + z) ^ 2 * (z + x) ^ 2)  :=  by sorry
Source
https://huggingface.co/datasets/internlm/Lean-Workbook/blob/main/lean_workbook.json, record lean_workbook_28399; Apache-2.0

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