lean_workbook_plus_17820

Proved

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

Let $a,b>0$ and $\frac{a^{2}}{1+b} + \frac{b^{2}}{1 + a}=\frac{8}{3}.$ Prove that $\frac{1}{a}+\frac{1}{b} \geq 1$

Preamble

import Mathlib.Analysis.Complex.Basic

Formal statement

theorem lean_workbook_plus_17820 (a b : ℝ) (ha : 0 < a) (hb : 0 < b) (hab : a * b ≠ 1) (hab2 : a^2 / (1 + b) + b^2 / (1 + a) = 8 / 3) : 1 / a + 1 / b ≥ 1   :=  by sorry

Source

https://huggingface.co/datasets/internlm/Lean-Workbook

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