lean_workbook_plus_56976

⚠️ 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.

Prove: \n\nSince $0\le x,y,z\le 1$ then \n\n $\frac{x}{{yz + 1}} + \frac{y}{{zx + 1}} + \frac{z}{{xy + 1}} \le \frac{x}{{xyz + 1}} + \frac{y}{{xyz+ 1}} + \frac{z}{{xyz + 1}}$ \n\nSo we just need show \n\n $\frac{x}{{xyz + 1}} + \frac{y}{{xyz+ 1}} + \frac{z}{{xyz + 1}}\le 2\Longleftrightarrow x+y+z\le 2xyz+2$ \n\nLet $f(x)=(2yz-1)x+2-y-z$ which is a linear function, so it remain to prove \n\n $f(1)\ge 0\Longleftrightarrow 2yz-y-z+1=yz+(y-1)(z-1)\ge 0$ \n\n $f(0)\ge 0\Longleftrightarrow 2-y-z\ge 0$ \n\nWhich is obviously true For $0\le y,z\le 1$

Why this node was retired

The posted statement is

theorem lean_workbook_plus_56976 : ∀ x y z : ℝ, (0 ≤ x ∧ 0 ≤ y ∧ 0 ≤ z ∧ x ≤ 1 ∧ y ≤ 1 ∧ z ≤ 1) → x / (y * z + 1) + y / (z * x + 1) + z / (x * y + 1) ≤ 2 ↔ x + y + z ≤ 2 * x * y * z + 2   :=  by sorry

Implication/iff scope lets the bound hypothesis apply only to the left side, so a point outside the unit cube refutes the equivalence.

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

Proposed corrected statement

Parenthesize the conclusion: for0≤x,y,z≤1 prove both the reciprocal bound≤2 and x+y+z≤2xyz+2, or their iff inside the implication. Do not place the domain restriction inside only one side of an iff.

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.