A shifted cyclic quadratic ratio lower bound at fixed sum three

Proved

WorkbookSource.base_48922

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

lean-workbooksource-checked

Given 3 positive real numbers a, b, c. That satisfy $a + b + c = 3$ . Prove that:

$\frac{3ab + 5a}{b^2+4b+3}+\frac{3bc + 5b}{c^2+4c+3}+\frac{3ca + 5c}{a^2+4a+3} \geq 3$

Source: InternLM Lean-Workbook, record lean_workbook_48922 (Apache-2.0). Complete source proposition preserved; proof developed independently.

Preamble

import Mathlib
open Real Nat

Formal statement

theorem WorkbookSource.base_48922 (a b c : ℝ) (habc : a + b + c = 3) (ha : 0 < a) (hb : 0 < b) (hc : 0 < c) : (3 * a * b + 5 * a) / (b ^ 2 + 4 * b + 3) + (3 * b * c + 5 * b) / (c ^ 2 + 4 * c + 3) + (3 * c * a + 5 * c) / (a ^ 2 + 4 * a + 3) ≥ 3  :=  by sorry

Source

https://huggingface.co/datasets/internlm/Lean-Workbook/blob/main/lean_workbook.json, record lean_workbook_48922; Apache-2.0

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