Prove2Me

A pairwise squared reciprocal product sum upper bound

Proved
WorkbookSource.plus_4054

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

lean-workbooksource-checked

Suppose that x,y,zx,y,z are positive real numbers. Prove that 1(x+y)2(y+z)2+1(y+z)2(z+x)2+1(z+x)2(x+y)2≤916xyz(x+y+z)\frac{1}{(x+y)^2(y+z)^2}+\frac{1}{(y+z)^2(z+x)^2}+\frac{1}{(z+x)^2(x+y)^2} \leq \frac{9}{16xyz(x+y+z)}

Source: InternLM Lean-Workbook, record lean_workbook_plus_4054 (Apache-2.0). Complete source proposition preserved; proof developed independently.

Preamble
import Mathlib
open Real Nat
Formal statement
theorem WorkbookSource.plus_4054 (x y z : ℝ) (hx : 0 < x) (hy : 0 < y) (hz : 0 < z) : (1 / (x + y) ^ 2 * 1 / (y + z) ^ 2 + 1 / (y + z) ^ 2 * 1 / (z + x) ^ 2 + 1 / (z + x) ^ 2 * 1 / (x + y) ^ 2) ≤ 9 / (16 * x * y * z * (x + y + z))   :=  by sorry
Source
https://huggingface.co/datasets/internlm/Lean-Workbook/blob/main/lean_workbook.json, record lean_workbook_plus_4054; Apache-2.0

View graph