lean_workbook_plus_1924

⚠️ 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.

Let $a\geq b\geq c.$ Then $\frac{a^2}{a^2+2bc}\geq\frac{b^2}{b^2+2ac}$ and $a-c\geq b-c.$ \n Thus, $\sum_{cyc}\frac{a^2(a-b)(a-c)}{a^2+2bc}\geq(a-b)(b-c)\left(\frac{a^2}{a^2+2bc}-\frac{b^2}{b^2+2ac}\right)+$ \n $+\frac{c^2(c-a)(c-b)}{c^2+2ab}\geq0.$

Why this node was retired

The posted statement is

theorem lean_workbook_plus_1924 :
  ∀ a b c : ℝ, a ≥ b ∧ b ≥ c → (a^2 / (a^2 + 2 * b * c) - b^2 / (b^2 + 2 * a * c)) * (a - b) + (c^2 / (c^2 + 2 * a * b) - a^2 / (a^2 + 2 * b * c)) * (b - c) + (b^2 / (b^2 + 2 * a * c) - c^2 / (c^2 + 2 * a * b)) * (c - a) ≥ 0   :=  by sorry

The target cyclic expression does not match the source weighted-square expression, and nonnegative/positive denominator conditions are absent.

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∑cyc a²(a−b)(a−c)/(a²+2bc) and the displayed lower decomposition, with a≥b≥c>0, then verify every algebraic identity.

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.