lean_workbook_plus_53528

⚠️ 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 $a, b, c$ are non-negative number which satisfying $a+b+c=3$ . Prove that:\n $\sqrt{a^2+b^2}+\sqrt[3]{b^3+c^3}+\sqrt[4]{c^4+a^4} \ge 6$ \nNQA.\nHappy new year, squing and xzlbq!

Why this node was retired

The posted statement is

theorem lean_workbook_plus_53528 (a b c : ℝ) (ha : a + b + c = 3) (ha' : a ≥ 0 ∧ b ≥ 0 ∧ c ≥ 0) : Real.sqrt (a ^ 2 + b ^ 2) + (b ^ 3 + c ^ 3) ^ (1 / 3) + (c ^ 4 + a ^ 4) ^ (1 / 4) ≥ 6   :=  by sorry

Two source roots use natural exponents; but the source bound itself is also false at a=b=c=1.

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

Confirmed directly from the Lean text: the numeric-literal exponents carry no type ascription, so they elaborate at type ℕ where division truncates (^ (1 / 3) → ^0; ^ (1 / 4) → ^0).

Proposed corrected statement

Use genuine cube/fourth roots and ask for the sharp minimum of the sum under nonnegative a+b+c=3; do not retain the unsupported lower bound 6.

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.