lean_workbook_plus_45354

⚠️ 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 $f={{(a^2+1)^2\over {b^2+c^2}}+{{(b^2+1)^2\over {a^2+c^2}}+{{(c^2+1)^2\over {a^2+b^2}}.}}}$ \nThen $[{(b^2+c^2)+(a^2+c^2)+(a^2+b^2)]\cdot f\ge [(a^2+1)+(b^2+1)+(c^2+1)]^2,}$ or \nf\ge{{(a^2+b^2+c^2+3)^2\over {2(a^2+b^2+c^2)}}={1\over 2}\left[ (a^2+b^2+c^2)+{9\over {a^2+b^2+c^2}}+6\right].\ (1)

Why this node was retired

The posted statement is

theorem lean_workbook_plus_45354 :  ∀ a b c : ℝ, (b^2 + c^2 + a^2 + c^2 + a^2 + b^2) * (a^2 + 1)^2 / (b^2 + c^2) + (b^2 + 1)^2 / (a^2 + c^2) + (c^2 + 1)^2 / (a^2 + b^2) ≥ (a^2 + 1 + b^2 + 1 + c^2 + 1)^2   :=  by sorry

The formal statement changes or incorrectly supplies part of the source problem; the accepted note identifies a concrete witness or definitional mechanism.

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 b²+c²,a²+c²,a²+b²>0, define f as the entire sum of three fractions, and prove 2(a²+b²+c²)*f≥(a²+b²+c²+3)². Restore parentheses so the prefactor multiplies all three summands.

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.