A sum-of-squares upper bound from a reciprocal difference constraint

Proved

WorkbookCorrected.plus_41062

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

corrected-formalizationlean-workbooksource-checked

Let $a,b,c\geq1$ and $a+b+c=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+6.$ Prove that $a^2+b^2+c^2\leq 21+6\sqrt{10}$

Formalization Note: The original formalization added abc=1, which is absent from the source. This correction removes that extra hypothesis and proves the source upper bound using a,b,c≥1 and the reciprocal relation alone.

Source: InternLM Lean-Workbook, record lean_workbook_plus_41062 (Apache-2.0).

Preamble

import Mathlib

Formal statement

theorem WorkbookCorrected.plus_41062 (a b c : ℝ) (ha : 1≤a) (hb : 1≤b) (hc : 1≤c)
    (h : a+b+c=1/a+1/b+1/c+6) : a^2+b^2+c^2 ≤ 21+6*Real.sqrt 10 := by sorry

Source

https://huggingface.co/datasets/internlm/Lean-Workbook/blob/main/lean_workbook.json, record lean_workbook_plus_41062; Apache-2.0

View graph