lean_workbook_plus_7760

Disproved

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

$(4a+4b+c)(\frac{a}{4}+\frac{b}{4}+c)\ge(a+b+c)^2\n$ \frac{a}{4a+4b+c}\le\frac{a(a+b+4c)}{4(a+b+c)^2}$

Preamble

import Mathlib.Analysis.Complex.Basic

Formal statement

theorem lean_workbook_plus_7760 : ∀ a b c : ℝ, (4 * a + 4 * b + c) * (a / 4 + b / 4 + c) ≥ (a + b + c) ^ 2   :=  by sorry

Source