lean_workbook_plus_13156

Proved

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

Let $a,\ b,\ c$ and $d$ be positive real numbers such that $a^2+b^2=c^2+d^2=1.$ Prove that $\frac{b}{a}+\frac{d}{c}\geq 2\frac{b+d}{a+c}.$ Proposed by A certain Japanese High School Student

Preamble

import Mathlib.Analysis.Complex.Basic

Formal statement

theorem lean_workbook_plus_13156 {a b c d : ℝ} (ha : 0 < a) (hb : 0 < b) (hc : 0 < c) (hd : 0 < d) (hab : a^2 + b^2 = 1) (hcd : c^2 + d^2 = 1) : b / a + d / c ≥ 2 * (b + d) / (a + c)   :=  by sorry

Source

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

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