lean_workbook_plus_61450

Proved

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

Let $a,b>0$ and $\dfrac{1}{a^2+1}+\dfrac{2}{b^2+1}=1.$ Prove that $a(4b-a) \leq 6$

Preamble

import Mathlib.Analysis.Complex.Basic

Formal statement

theorem lean_workbook_plus_61450 (a b : ℝ) (ha : 0 < a) (hb : 0 < b) (hab : 1 / (a^2 + 1) + 2 / (b^2 + 1) = 1) : a * (4 * b - a) ≤ 6   :=  by sorry

Source