A reciprocal sum times a shifted square has a linear lower bound

Proved

WorkbookSource.base_49195

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

lean-workbooksource-checked

Prove that $(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})(a+b+c-1)^2+12 \ge 8(a+b+c)$ given $a,b,c > 0$ .

Source: InternLM Lean-Workbook, record lean_workbook_49195 (Apache-2.0). Complete source proposition preserved; proof developed independently.

Preamble

import Mathlib
open Real Nat

Formal statement

theorem WorkbookSource.base_49195 (a b c : ℝ) (ha : a > 0) (hb : b > 0) (hc : c > 0) : (1 / a + 1 / b + 1 / c) * (a + b + c - 1) ^ 2 + 12 ≥ 8 * (a + b + c)  :=  by sorry

Source

https://huggingface.co/datasets/internlm/Lean-Workbook/blob/main/lean_workbook.json, record lean_workbook_49195; Apache-2.0

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