lean_workbook_plus_42708

Proved

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

Let $a,b,c\geq 0$ . Prove that \n\n

a^2+2b^2+c^2 \ge \frac{2\sqrt{3}}{3}(a+2b)c

Preamble

import Mathlib.Analysis.Complex.Basic

Formal statement

theorem lean_workbook_plus_42708 : ∀ a b c : ℝ, a ≥ 0 ∧ b ≥ 0 ∧ c ≥ 0 → a^2 + 2 * b^2 + c^2 ≥ (2 * Real.sqrt 3) / 3 * (a + 2 * b) * c   :=  by sorry

Source