lean_workbook_plus_82364

⚠️ 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 $a,b,c\geq 0$ and satisfying $a^{2}+b^{2}+c^{2}+2abc=1$ Prove $a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}+2(abc)^{\frac{5}{3}}\geq 2abc$

Why this node was retired

The posted statement is

theorem lean_workbook_plus_82364 (a b c : ℝ) (ha : a ≥ 0) (hb : b ≥ 0) (hc : c ≥ 0) (habc : a * b * c ≠ 0) (h : a^2 + b^2 + c^2 + 2 * a * b * c = 1) :
  a^2 * b^2 + b^2 * c^2 + c^2 * a^2 + 2 * (a * b * c)^(5 / 3) ≥ 2 * a * b * c   :=  by sorry

The exponent truncates to1, so the two2abc terms cancel and the remaining squared terms are nonnegative.

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 (^(5 / 3) → ^1).

Proposed corrected statement

Use the real exponent5/3 and remove the extra abc≠0 hypothesis absent from the nonnegative source.

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.