lean_workbook_plus_50970

⚠️ 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.

If $t\geq0$, then\n$\left( {{\it x_1}}^{2} + {{\it x_2}}^{2} + {{\it x_3}}^{2} + {{\it x_4}}^{2} \right) \left( {{\it x_1}}^{ - 2} + {{\it x_2}}^{ - 2} + {{\it x_3}}^{ - 2} + {{\it x_4}}^{ - 2} \right) + 2\,t \geq 2\,{\frac {\left( \left( {\it x_1} + {\it x_2} + {\it x_3} + {\it x_4} \right) \left( {{\it x_1}}^{ - 1} + {{\it x_2}}^{ - 1} + {{\it x_3}}^{ - 1} + {{\it x_4}}^{ - 1} \right) - 8 + t \right) ^{2}}{8 + t}}$.

Why this node was retired

The posted statement is

theorem lean_workbook_plus_50970 :  ∀ x1 x2 x3 x4 t : ℝ, x1 ≥ 0 ∧ x2 ≥ 0 ∧ x3 ≥ 0 ∧ x4 ≥ 0 ∧ t ≥ 0 →   (x1^2 + x2^2 + x3^2 + x4^2) * (x1^(-2:ℤ) + x2^(-2:ℤ) + x3^(-2:ℤ) + x4^(-2:ℤ)) + 2 * t ≥   2 * (x1 + x2 + x3 + x4) * (x1^(-1:ℤ) + x2^(-1:ℤ) + x3^(-1:ℤ) + x4^(-1:ℤ)) - 8 + t   :=  by sorry

The right-hand square and denominator8+t were omitted; nonzero inverse inputs are not enforced.

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

Proposed corrected statement

For x₁,…,x₄>0 and t≥0 restore (∑xᵢ²)(∑xᵢ^(−2))+2t≥2((∑xᵢ)(∑xᵢ^(−1))−8+t)²/(8+t), with the entire squared numerator grouped.

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.