lean_workbook_plus_1800

⚠️ 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.

We claim that $\frac{1}{\sqrt{\frac{1}{2}+p+pq+pqr}} \ge \frac{\sqrt{2}}{1+p+pq+pqr}$ . Indeed, setting $x=p,y=pq,z=pqr$ , it becomes \n \begin{align*}& \frac{1}{\sqrt{\frac 12+x+y+z}}\ge \frac{\sqrt{2}}{1+x+y+z}\n\iff & (1+x+y+z)^2\ge 2\left( \frac 12+x+y+z\right)\n\iff & x^2+y^2+z^2+2xy+2yz+2zx\ge 0\n\iff & (x+y+z)^2\ge 0\end{align*} The last inequality is obvious.

Why this node was retired

The posted statement is

theorem lean_workbook_plus_1800 :
  ∀ p q r : ℝ, (1 / Real.sqrt ((1:ℝ) / 2 + p + p * q + p * q * r)) ≥ Real.sqrt 2 / (1 + p + p * q + p * q * r)   :=  by sorry

The radical denominator can be zero; the source squaring/cancellation argument requires its positive domain.

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

Proposed corrected statement

Writing s=p+pq+pqr, assume s>−1/2 and prove 1/√(1/2+s)≥√2/(1+s); both denominators are then positive and the comparison reduces to s²≥0.

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.