lean_workbook_plus_26361

Proved

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

With x,y,z>0 we have $\frac{a^3}{x}+\frac{b^3}{y}+\frac{c^3}{z}\geq\frac{a^3+b^3+c^3}{3(x+y+z)}$

Preamble

import Mathlib.Analysis.Complex.Basic

Formal statement

theorem lean_workbook_plus_26361 (a b c x y z : ℝ) (ha : 0 < a) (hb : 0 < b) (hc : 0 < c) (hx : 0 < x) (hy : 0 < y) (hz : 0 < z) : (a^3 / x + b^3 / y + c^3 / z) ≥ (a^3 + b^3 + c^3) / (3 * (x + y + z))   :=  by sorry

Source

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

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