A shifted cyclic reciprocal sum lower bound at fixed sum six

Proved

WorkbookSource.plus_40566

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

lean-workbooksource-checked

Given $a,b,c$ are positive real numbers satisfying $a+b+c =6$ . Prove that: $\frac{a}{b^2 + c + 1} + \frac{b}{c^2 + a + 1} + \frac{c}{a^2 + b + 1} \ge \frac{6}{7}$

Source: InternLM Lean-Workbook, record lean_workbook_plus_40566 (Apache-2.0). Complete source proposition preserved; proof developed independently.

Preamble

import Mathlib
open Real Nat

Formal statement

theorem WorkbookSource.plus_40566 (a b c : ℝ) (ha : 0 < a) (hb : 0 < b) (hc : 0 < c) (habc : a + b + c = 6) : a / (b ^ 2 + c + 1) + b / (c ^ 2 + a + 1) + c / (a ^ 2 + b + 1) ≥ 6 / 7   :=  by sorry

Source

https://huggingface.co/datasets/internlm/Lean-Workbook/blob/main/lean_workbook.json, record lean_workbook_plus_40566; Apache-2.0

View graph