lean_workbook_plus_53550

⚠️ 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 $a$ and $b$ positive real numbers such that $a^2+b^2=3$ and $2a\geq b$ , prove that: $\frac{6ab^2-3b^3+b^2}{a^2} \leq 14-6\sqrt{2}.$ Equality holds when $b=\sqrt{2}a=\sqrt{2}.$

Why this node was retired

The posted statement is

theorem lean_workbook_plus_53550 : ∀ a b : ℝ, a^2 + b^2 = 3 ∧ 2*a ≥ b → (6*a*b^2 - 3*b^3 + b^2)/a^2 ≤ 14 - 6 * Real.sqrt 2 ∧ (b = Real.sqrt 2 * a ∧ b = Real.sqrt 2)   :=  by sorry

The equality-case values are asserted for every admissible input, and source positivity is omitted.

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

Proposed corrected statement

Require a,b>0,a²+b²=3,2a≥b. Prove the inequality, and separately state equality iff a=1,b=√2; do not conjoin that equality case to every conclusion.

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.