lean_workbook_plus_69488

⚠️ Retired — specification defect

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

Let $a,b,c\ge 0$ and $a^2b+b^2c+c^2a+abc=\frac{1}{3}$ . Prove that

Why this node was retired

The posted statement is

theorem lean_workbook_plus_69488 (a b c : ℝ) (ha : a ≥ 0) (hb : b ≥ 0) (hc : c ≥ 0) (habc : a * b * c = 1 / 3) (h : a^2 * b + b^2 * c + c^2 * a + a * b * c = 1 / 3) : a + b + c ≥ (9 / 4)^(1 / 3)   :=  by sorry

The added abc=1/3 contradicts the other nonnegative cubic sum equaling1/3; the cube-root exponent also becomes0.

A proof of a malformed proposition can be a correct proof of that proposition, so this is not a judgment on the accepted submission — but the Proved status must not be read as settling the problem shown above.

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

Remove the extra normalization; retain exactly the positivity/domain restrictions and the displayed source constraint, then prove the displayed source conclusion. Source task: Let $a,b,c\ge 0$ and $a^2b+b^2c+c^2a+abc=\frac{1}{3}$ . Prove that

Use the genuine nonnegative cube root of9/4.

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.