A reciprocal lower bound from a quadratic denominator identity

Proved

WorkbookCorrected.plus_35576

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

corrected-formalizationlean-workbooksource-checked

Let $a,b>0$ satisfy

\frac1{a^2+2}+\frac1{b^2+2}=\frac13.

Then $\frac1a+\frac1b\ge1$ .

Formalization Note: Parentheses are restored in the two denominators, matching the source. The original formalization parsed the added2 outside each fraction, producing inconsistent assumptions.

Source: InternLM Lean-Workbook, record lean_workbook_plus_35576 (Apache-2.0).

Preamble

import Mathlib

Formal statement

theorem WorkbookCorrected.plus_35576 (a b : ℝ) (ha : 0<a) (hb : 0<b) (h : 1/(a^2+2)+1/(b^2+2)=(1/3 : ℝ)) : 1/a+1/b ≥ 1 := by sorry

Source

https://huggingface.co/datasets/internlm/Lean-Workbook/blob/main/lean_workbook.json, record lean_workbook_plus_35576; Apache-2.0

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