A shifted cyclic quadratic ratio lower bound at fixed sum three

Proved

WorkbookSource.base_20698

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

lean-workbooksource-checked

Let $a$ , $b$ , $c$ be positive real numbers such that $a+b+c=3$ . Prove that $\frac{3a+1}{3b^2+1}+\frac{3b+1}{3c^2+1}+\frac{3c+1}{3a^2+1}\geq 3.$

Source: InternLM Lean-Workbook, record lean_workbook_20698 (Apache-2.0). Complete source proposition preserved; proof developed independently.

Preamble

import Mathlib
open Real Nat

Formal statement

theorem WorkbookSource.base_20698 (a b c : ℝ) (ha : a > 0) (hb : b > 0) (hc : c > 0) (hab : a + b + c = 3) : (3 * a + 1) / (3 * b ^ 2 + 1) + (3 * b + 1) / (3 * c ^ 2 + 1) + (3 * c + 1) / (3 * a ^ 2 + 1) ≥ 3  :=  by sorry

Source

https://huggingface.co/datasets/internlm/Lean-Workbook/blob/main/lean_workbook.json, record lean_workbook_20698; Apache-2.0

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