lean_workbook_plus_63792

Proved

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

Let $a,b,c>0$ such that $\frac{a}{b+c}+\frac{b}{c+a}\leq\frac{3}{2}.$ Then $\frac{c}{a+b}\ge \frac{1}{6}.$

Preamble

import Mathlib.Analysis.Complex.Basic

Formal statement

theorem lean_workbook_plus_63792 (a b c : ℝ) (ha : 0 < a) (hb : 0 < b) (hc : 0 < c) : (a / (b + c) + b / (c + a) ≤ 3 / 2 → c / (a + b) ≥ 1 / 6)   :=  by sorry

Source

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

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