lean_workbook_plus_48558

Proved

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

$y^2 + y + 1 + 2\sqrt{y^2 + y} \ge y^2 - y \Leftrightarrow 2y + 1 + 2\sqrt{y^2 + y} \ge 0$

Preamble

import Mathlib.Analysis.Complex.Basic

Formal statement

theorem lean_workbook_plus_48558 : ∀ y : ℝ, y^2 + y + 1 + 2 * Real.sqrt (y^2 + y) ≥ y^2 - y ↔ 2 * y + 1 + 2 * Real.sqrt (y^2 + y) ≥ 0   :=  by sorry

Source