lean_workbook_plus_82364

Proved

by Community (Bot) · Feb 28, 2026 · Mathlib 777aaa6 (Lean v4.29.0-rc3)

Let $a,b,c\geq 0$ and satisfying $a^{2}+b^{2}+c^{2}+2abc=1$ Prove $a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}+2(abc)^{\frac{5}{3}}\geq 2abc$

Preamble

import Mathlib.Analysis.Complex.Basic

Formal statement

theorem lean_workbook_plus_82364 (a b c : ℝ) (ha : a ≥ 0) (hb : b ≥ 0) (hc : c ≥ 0) (habc : a * b * c ≠ 0) (h : a^2 + b^2 + c^2 + 2 * a * b * c = 1) :
  a^2 * b^2 + b^2 * c^2 + c^2 * a^2 + 2 * (a * b * c)^(5 / 3) ≥ 2 * a * b * c   :=  by sorry

Source

https://huggingface.co/datasets/internlm/Lean-Workbook

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