Prove2Me

A shifted cubic ratio lower bound at fixed sum three

Proved
WorkbookSource.base_17772

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

lean-workbooksource-checked

Show that , a+b3c+1+b+c3a+1+c+a3b+1≥3\frac{a+b^3}{c+1}+\frac{b+c^3}{a+1}+\frac{c+a^3}{b+1}\ge 3, where a+b+c=3,a,b,c∈R+a+b+c=3,a,b,c\in \mathbb{R}^+.

Source: InternLM Lean-Workbook, record lean_workbook_17772 (Apache-2.0). Complete source proposition preserved; proof developed independently.

Preamble
import Mathlib
open Real Nat
Formal statement
theorem WorkbookSource.base_17772 (a b c : ℝ) (ha : 0 < a) (hb : 0 < b) (hc : 0 < c) (habc : a + b + c = 3) : (a + b^3) / (c + 1) + (b + c^3) / (a + 1) + (c + a^3) / (b + 1) ≥ 3  :=  by sorry
Source
https://huggingface.co/datasets/internlm/Lean-Workbook/blob/main/lean_workbook.json, record lean_workbook_17772; Apache-2.0

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