lean_workbook_plus_47535

⚠️ 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.

From the given equations we get $xz+y=2z$ $yx+z=2x$ $zy+x=2y$ Adding them yields $xy+xz+yz = x+y+z$ Suppose wlog $x \ge y \ge z$ . Hence $xy \ge xz \ge yz$ . If $xy < z => xy+xz+yz < 3z \le x+y+z$ . Hence $xy \ge z => x\ge \frac {z}{y}$ . Therefore we have $2 = x+ \frac {y}{z} \ge \frac{z}{y} + \frac{y}{z} \ge 2 => y=z$ Finally $x+1 =2 => x=1$ and from $y+z=2$ and $z+\frac{1}{y} = 2$ we get $y=z=1$ . Therefore $x+y+z = 3$

Why this node was retired

The posted statement is

theorem lean_workbook_plus_47535  (x y z : ℝ)
  (h₀ : x + y + z = 2 * x)
  (h₁ : y + z = 2 * y)
  (h₂ : z + x = 2 * z) :
  x + y + z = 3   :=  by sorry

The three source product equations are replaced by different linear equations and the positive-variable setting of divisions is lost.

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

Proposed corrected statement

Restore xz+y=2z,xy+z=2x,yz+x=2y with x,y,z>0 before deriving x=y=z=1.

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.