lean_workbook_plus_32246

Proved

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

Or, by AM-GM, $a^{100} + b^{100} \geq 2 \sqrt{a^{100} \cdot b^{100}} = 2(ab)^{50}$ .

Preamble

import Mathlib.Analysis.Complex.Basic

Formal statement

theorem lean_workbook_plus_32246  (a b : ℝ)
  (h₀ : 0 < a ∧ 0 < b) :
  a^100 + b^100 ≥ 2 * (a * b)^50   :=  by sorry

Source