A shifted reciprocal sum has a reciprocal product upper bound

Proved

WorkbookSource.plus_25937

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

lean-workbooksource-checked

Let $a,b,c$ are positive numbers satisfies $a+b+c=3$ , prove that $\frac{1}{2a+bc}+\frac{1}{2b+ca}+\frac{1}{2c+ab} \leq \frac{1}{abc}.$

Source: InternLM Lean-Workbook, record lean_workbook_plus_25937 (Apache-2.0). Complete source proposition preserved; proof developed independently.

Preamble

import Mathlib
open Real Nat

Formal statement

theorem WorkbookSource.plus_25937 (a b c : ℝ) (ha : 0 < a) (hb : 0 < b) (hc : 0 < c) (habc : a + b + c = 3) : 1 / (2 * a + b * c) + 1 / (2 * b + c * a) + 1 / (2 * c + a * b) ≤ 1 / (a * b * c)   :=  by sorry

Source

https://huggingface.co/datasets/internlm/Lean-Workbook/blob/main/lean_workbook.json, record lean_workbook_plus_25937; Apache-2.0

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