lean_workbook_plus_63806

⚠️ Retired — mistyped fractional exponent

The exponent ^(1 / 5) in the Lean statement below is natural-number division, not a real exponent, so it collapses 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.

$\dfrac{1}{1+c} \geq 5\sqrt[5]{\dfrac{ab^2c^2}{(1+a)(1+b)^2(1+c)^2}}$

Why this node was retired

The posted statement is

theorem lean_workbook_plus_63806 : ∀ a b c : ℝ, a > 0 ∧ b > 0 ∧ c > 0 → 1 / (1 + c) ≥ 5 * (a * b^2 * c^2 / ((1 + a) * (1 + b)^2 * (1 + c)^2))^(1 / 5)   :=  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:

• ^(1 / 5) → ^0, so the radical is 1 for every base

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

Use genuine real roots and explicitly real fractional exponents. For odd roots of negative values use the signed real root. Retain the source domain and verify the repaired claim rather than infer its truth from this counterexample. Original task: $\dfrac{1}{1+c} \geq 5\sqrt[5]{\dfrac{ab^2c^2}{(1+a)(1+b)^2(1+c)^2}}$ Recover the actual source relation on a,b,c; restoring the fifth root alone does not make the printed bound true.

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.