A shifted quadratic reciprocal sum upper bound at fixed sum three

Proved

WorkbookSource.plus_26490

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

lean-workbooksource-checked

If $a, b, c>0, a+b+c=3$ prove that $\frac{1}{a^2+bc+1}+\frac{1}{b^2+ca+1}+\frac{1}{c^2+ab+1}\le\frac{a^2+b^2+c^2}{3abc}$

Source: InternLM Lean-Workbook, record lean_workbook_plus_26490 (Apache-2.0). Complete source proposition preserved; proof developed independently.

Preamble

import Mathlib
open Real Nat

Formal statement

theorem WorkbookSource.plus_26490 (a b c : ℝ) (ha : 0 < a) (hb : 0 < b) (hc : 0 < c) (hab : a + b + c = 3) : 1 / (a^2 + b * c + 1) + 1 / (b^2 + c * a + 1) + 1 / (c^2 + a * b + 1) ≤ (a^2 + b^2 + c^2) / (3 * a * b * c)   :=  by sorry

Source

https://huggingface.co/datasets/internlm/Lean-Workbook/blob/main/lean_workbook.json, record lean_workbook_plus_26490; Apache-2.0

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