A weighted triple-sum ratio sum is at least sixteen thirds

Proved

WorkbookSource.base_6786

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

lean-workbooksource-checked

Prove or disprove that for $a, b, c, d > 0$ and $abc = 1$ , $\frac{2a+b+d}{a+c+d}+\frac{2b+c+a}{b+d+a}+\frac{2c+d+b}{c+a+b}+\frac{2d+a+c}{d+b+c}\geq\frac{16}{3}$

Source: InternLM Lean-Workbook, record lean_workbook_6786 (Apache-2.0). Complete source proposition preserved; proof developed independently.

Preamble

import Mathlib
open Real Nat

Formal statement

theorem WorkbookSource.base_6786 (a b c d : ℝ) (ha : 0 < a) (hb : 0 < b) (hc : 0 < c) (hd : 0 < d) (habc : a * b * c = 1) : (2 * a + b + d) / (a + c + d) + (2 * b + c + a) / (b + d + a) + (2 * c + d + b) / (c + a + b) + (2 * d + a + c) / (d + b + c) ≥ 16 / 3  :=  by sorry

Source

https://huggingface.co/datasets/internlm/Lean-Workbook/blob/main/lean_workbook.json, record lean_workbook_6786; Apache-2.0

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