lean_workbook_plus_5152

⚠️ 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.

Prove that : $\sqrt[3]{\frac{a}{b+c}}+\sqrt[3]{\frac{b}{c+a}}+\sqrt[3]{\frac{c}{a+b}}\geq 2$

Why this node was retired

The posted statement is

theorem lean_workbook_plus_5152 (a b c : ℝ) : (a / (b + c))^(1 / 3) + (b / (c + a))^(1 / 3) + (c / (a + b))^(1 / 3) ≥ 2   :=  by sorry

Natural cube-root exponents give3≥2. The source fragment and target have no sign/nonzero conditions, so the genuine expression is undefined at some inputs and false under totalization at a=b=c=0.

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

For a,b,c≥0 with ab+bc+ca>0, prove∛(a/(b+c))+∛(b/(c+a))+∛(c/(a+b))≥2 using genuine nonnegative roots. This supplies explicit positive denominators; confirm the original domain from the full 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.