lean_workbook_plus_24610

⚠️ 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.

$\frac{a^6}{a^4+a^3b+a^2b^2}+\frac{b^6}{b^4+b^3c+b^2c^2}+\frac{c^6}{c^4+ac^3+c^2a^2} \geq \frac{(a^3+b^3+c^3)^2}{a^4+b^4+c^4+a^3b+b^3c+c^3a+a^2b^2+b^2c^2+c^2a^2} \geq \frac{(a^3+b^3+c^3)^2}{(a^3+b^3+c^3)(a+b+c)} = a^3+b^3+c^3$

Why this node was retired

The posted statement is

theorem lean_workbook_plus_24610 : ∀ a b c : ℝ, (a ^ 6 / (a ^ 4 + a ^ 3 * b + a ^ 2 * b ^ 2) + b ^ 6 / (b ^ 4 + b ^ 3 * c + b ^ 2 * c ^ 2) + c ^ 6 / (c ^ 4 + c ^ 3 * a + c ^ 2 * a ^ 2) ≥ (a ^ 3 + b ^ 3 + c ^ 3) ^ 2 / (a ^ 4 + b ^ 4 + c ^ 4 + a ^ 3 * b + b ^ 3 * c + c ^ 3 * a + a ^ 2 * b ^ 2 + b ^ 2 * c ^ 2 + c ^ 2 * a ^ 2)) ∧ (a ^ 3 + b ^ 3 + c ^ 3) ^ 2 / (a ^ 4 + b ^ 4 + c ^ 4 + a ^ 3 * b + b ^ 3 * c + c ^ 3 * a + a ^ 2 * b ^ 2 + b ^ 2 * c ^ 2 + c ^ 2 * a ^ 2) >= (a ^ 3 + b ^ 3 + c ^ 3) ^ 2 / (a ^ 3 + b ^ 3 + c ^ 3) * (a + b + c)   :=  by sorry

The final source denominator is a product, whereas the target divides by the cube sum then multiplies by a+b+c. The source final simplification also omits this latter denominator.

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

Proposed corrected statement

Restore (a³+b³+c³)²/((a³+b³+c³)(a+b+c))=(a³+b³+c³)/(a+b+c), with the ordinary positive domain; recover any normalization a+b+c=1 if that was intended.

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.