lean_workbook_plus_74162

Proved

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

prove that $\sqrt{1+a^2}\le \frac{(a+b)+(a+c)}{2}$ given $a,b,c>0$ and $ab+bc+ca=1$

Preamble

import Mathlib.Analysis.Complex.Basic

Formal statement

theorem lean_workbook_plus_74162 (a b c : ℝ) (ha : 0 < a) (hb : 0 < b) (hc : 0 < c) (habc : a * b + b * c + c * a = 1) : Real.sqrt (1 + a ^ 2) ≤ (a + b + (a + c)) / 2   :=  by sorry

Source

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

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