lean_workbook_plus_67741

⚠️ 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.

For $a,b,c > 0$ and $ab + bc + ca = 3$ prove that : $3 + \frac {(a - b)^2 + (b - c)^2 + (c - a)^2}{2} \ge \frac {a + b^2c^2}{b + c} + \frac {b + c^2a^2}{c + a} + \frac {c + a^2b^2}{a + b}\ge 3$

Why this node was retired

The posted statement is

theorem lean_workbook_plus_67741 :  ∀ a b c : ℝ, a > 0 ∧ b > 0 ∧ c > 0 ∧ a * b + b * c + c * a = 3 → 3 + (a - b) ^ 2 / 2 + (b - c) ^ 2 / 2 + (c - a) ^ 2 / 2 ≥ a + b ^ 2 * c ^ 2 / (b + c) + b + c ^ 2 * a ^ 2 / (c + a) + c + a ^ 2 * b ^ 2 / (a + b) ∧ a + b ^ 2 * c ^ 2 / (b + c) + b + c ^ 2 * a ^ 2 / (c + a) + c + a ^ 2 * b ^ 2 / (a + b) ≥ 3   :=  by sorry

Each source numerator is a+b²c², whereas division applies only to b²c² in the target.

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 (a+b²c²)/(b+c) and the cyclic variants in both inequalities.

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.