lean_workbook_plus_36624

⚠️ 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.

By Holder and AM-GM we obtain: $(7(2a+2b+c))^3\left(\frac{4a^3b}{(2a+3b)^3}+\frac{b^3c}{(2b+3c)^3}+\frac{2c^3a}{(2c+3a)^3}\right)=$ $(4(2a+3b)+(2b+3c)+2(2c+3a))^3\left(\frac{4a^3b}{(2a+3b)^3}+\frac{b^3c}{(2b+3c)^3}+\frac{2c^3a}{(2c+3a)^3}\right)\geq$ $\geq(4\sqrt[4]{a^3b}+\sqrt[4]{b^3c}+2\sqrt[4]{c^3a})^4\geq2401\sqrt[7]{a^{12}b^4\cdot b^3c\cdot c^6a^2}=2401a^2bc,$ which says $\sum_{cyc}\left(\frac{4a^3b}{(2a+3b)^3}+\frac{b^3c}{(2b+3c)^3}+\frac{2c^3a}{(2c+3a)^3}\right)\geq7\sum_{cyc}\frac{a^2bc}{(2a+2b+c)^3}$ or $\sum_{cyc}\frac{a^3b}{(2a+3b)^3}\geq\sum_{cyc}\frac{a^2bc}{(2a+2b+c)^3}$ and we are done!

Why this node was retired

The posted statement is

theorem lean_workbook_plus_36624 :
  ∀ a b c : ℝ, (4 * a^3 * b / (2 * a + 3 * b)^3 + 4 * b^3 * c / (2 * b + 3 * c)^3 + 2 * c^3 * a / (2 * c + 3 * a)^3) ≥
    7 * (a^2 * b * c / (2 * a + 2 * b + c)^3 + b^2 * c * a / (2 * b + 2 * c + a)^3 + c^2 * a * b / (2 * c + 2 * a + b)^3)   :=  by sorry

The target mixes coefficients from one weighted intermediate inequality with the final cyclic sum.

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 the cyclicly summed conclusion Σcyc a³b/(2a+3b)³≥Σcyc a²bc/(2a+2b+c)³ for positive a,b,c.

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.