lean_workbook_plus_32528

Proved

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

Prove the inequality: $\sqrt{\frac{a^2+b^2}{2}} \ge \dfrac{a+b}{2}$ for positive numbers $a, b$ .

Preamble

import Mathlib.Analysis.Complex.Basic

Formal statement

theorem lean_workbook_plus_32528 (a b : ℝ) (ha : 0 < a) (hb : 0 < b) :  Real.sqrt ((a^2 + b^2) / 2) ≥ (a + b) / 2   :=  by sorry

Source