lean_workbook_plus_59156

Proved

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

Given $a, b$ are positive and $a^{2}+b^{2}=1$ , prove that $1<a+b\leq{\sqrt{2}}$

Preamble

import Mathlib.Analysis.Complex.Basic

Formal statement

theorem lean_workbook_plus_59156 : ∀ a b : ℝ, a > 0 ∧ b > 0 ∧ a^2 + b^2 = 1 → 1 < a + b ∧ a + b ≤ Real.sqrt 2   :=  by sorry

Source