A reciprocal-sum comparison with cyclic weighted denominators

Proved

WorkbookSource.base_3999

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

lean-workbooksource-checked

Prove that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+3(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a})\ge 10(\frac{1}{3a+b}+\frac{1}{3b+c}+\frac{1}{3c+a})$ given $a, b, c > 0$ .

Source: InternLM Lean-Workbook, record lean_workbook_3999 (Apache-2.0). Complete source proposition preserved; proof developed independently.

Preamble

import Mathlib
open Real Nat

Formal statement

theorem WorkbookSource.base_3999 (a b c : ℝ) (ha : 0 < a) (hb : 0 < b) (hc : 0 < c) : (1 / a + 1 / b + 1 / c + 3 * (1 / (a + b) + 1 / (b + c) + 1 / (c + a))) ≥ 10 * (1 / (3 * a + b) + 1 / (3 * b + c) + 1 / (3 * c + a))  :=  by sorry

Source

https://huggingface.co/datasets/internlm/Lean-Workbook/blob/main/lean_workbook.json, record lean_workbook_3999; Apache-2.0

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