lean_workbook_plus_21265

⚠️ Retired — specification defect

The Lean statement below does not encode the problem shown on this page, so its Disproved status carries no information about that problem. Do not import this node or use it as a dependency.

prove that: $\sqrt [3]{6a + \frac {1}{c}} + \sqrt [3]{6b + \frac {1}{a}} + \sqrt [3]{6c + \frac {1}{b}} \le 1$, given $a,b,c \ge 0$ and $ab + ac + bc = 1$

Why this node was retired

The posted statement is

theorem lean_workbook_plus_21265 (a b c : ℝ) (ha : a ≥ 0) (hb : b ≥ 0) (hc : c ≥ 0) (hab : a * b + a * c + b * c = 1) : (6 * a + 1 / c)^(1 / 3) + (6 * b + 1 / a)^(1 / 3) + (6 * c + 1 / b)^(1 / 3) ≤ 1   :=  by sorry

Fractional-looking exponents elaborate as natural-number quotients, changing roots to powers0 (or another integer). The genuine-root source is also false for equal positive variables satisfying ab+ac+bc=1.

The recorded counterexample refutes the statement as encoded. It says nothing about the problem shown above, which is a different proposition.

Confirmed directly from the Lean text: the numeric-literal exponent carries no type ascription, so it elaborates at type ℕ where division truncates (^(1 / 3) → ^0).

Proposed corrected statement

Recover the source expression/sign/constant: with a=b=c=1/√3, the sum of the genuine cube roots of6a+1/c is far greater than1. Neither real typing nor excluding zero denominators repairs this bound.

Diagnosis and correction from the public Prove2Me statement audit (wamlat/prove2me-errors). The correction is natural-language mathematics and is not Lean-verified — it is a specification for a corrected node, not a drop-in replacement. No corrected replacement node exists yet.