lean_workbook_plus_79584

Proved

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

Let $a,b$ be positive real numbers such that $\frac{1}{a}+\frac{1}{b}=1.$ Prove that $6ab+\frac{1}{a+b}\leq \frac{65}{4}+a^2+b^2$

Preamble

import Mathlib.Analysis.Complex.Basic

Formal statement

theorem lean_workbook_plus_79584 (a b : ℝ) (ha : 0 < a) (hb : 0 < b) (hab : 1/a + 1/b = 1) : 6*a*b + 1/(a + b) ≤ 65/4 + a^2 + b^2   :=  by sorry

Source