lean_workbook_plus_44935

Proved

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

Let $a,b,c$ be non negative real numbers satisfying $a+b^2+c=1$ . Prove that $(1+a^2)(1+b^2)(1+c^2)\geq \frac{25}{16}$ Equality holds when $a=c=\frac{1}{2},b=0.$

Preamble

import Mathlib.Analysis.Complex.Basic

Formal statement

theorem lean_workbook_plus_44935 (a b c : ℝ) (ha : a ≥ 0) (hb : b ≥ 0) (hc : c ≥ 0) (hab : a + b^2 + c = 1) : (1 + a^2) * (1 + b^2) * (1 + c^2) ≥ 25 / 16 ∧ (a = 1 / 2 ∧ b = 0 ∧ c = 1 / 2) → (1 + a^2) * (1 + b^2) * (1 + c^2) = 25 / 16   :=  by sorry

Source

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

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