lean_workbook_plus_39159
Disproved⚠️ Retired — mistyped fractional exponent
The exponents
^ (2 / 3),^ (5 / 3)in the Lean statement below are natural-number division, not a real exponent, so they collapse to an integer power and the displayed radical is not what is being asserted. Do not import this node or use it as a dependency.
You can convert Muirhead into multiple applications of AM-GM, in the above case this is, \n\nBy AM-GM, \n\n
Why this node was retired
The posted statement is
theorem lean_workbook_plus_39159 : ∀ a b c : ℝ, a > 0 ∧ b > 0 ∧ c > 0 → 1 / 6 * (a ^ 2 * b + a ^ 2 * c + b ^ 2 * a + b ^ 2 * c + c ^ 2 * a + c ^ 2 * b) ≥ a ^ (5 / 3) * b ^ (2 / 3) * c ^ (2 / 3) := by sorry
A numeric-literal exponent carrying no type ascription is elaborated at type ℕ (via HPow ℝ ℕ ℝ, Monoid.npow), and natural-number division truncates. Each occurrence therefore collapses:
^ (2 / 3)→^0, so the radical is1for every base^ (5 / 3)→^1, i.e. the 5/3 power is replaced by the 1th power
The recorded counterexample refutes the collapsed statement. It says nothing about the exercise displayed on this page, which is a different proposition.
What a faithful statement would require
Every fractional exponent must be given a real type, e.g. x ^ ((1 : ℝ) / 3) using Real.rpow, or be written with the intended root operation (Real.sqrt, or the signed real cube root where the radicand may be negative). Real powers also need their own domain hypotheses: rpow is only the intended root for a nonnegative base.
Proposed corrected statement
For positive a,b,c apply AM–GM to a²b,a²c,ab²,ac²,a²b,a²c. Their arithmetic mean is at least a^(5/3)b^(2/3)c^(2/3) with real exponents. Recover the intended subsequent symmetric summation and normalization.
From the public Prove2Me statement audit (wamlat/prove2me-errors). That proposal is natural-language mathematics and is not Lean-verified; it is a specification for a corrected node, not a drop-in replacement.
import Mathlib.Analysis.Complex.Basic
theorem lean_workbook_plus_39159 : ∀ a b c : ℝ, a > 0 ∧ b > 0 ∧ c > 0 → 1 / 6 * (a ^ 2 * b + a ^ 2 * c + b ^ 2 * a + b ^ 2 * c + c ^ 2 * a + c ^ 2 * b) ≥ a ^ (5 / 3) * b ^ (2 / 3) * c ^ (2 / 3) := by sorry