ALG-B2-M02-P023 Two cyclic sums

Hint 1. Subtract the first equality from the second.

Hint 2. Obtain \((a-b)(b-c)(c-a)=2\), then denote two adjacent differences.

Subtracting gives \((a-b)(b-c)(c-a)=2\). Let \(c\) be the largest number. Then the signs imply \(a0\), \(y=c-b>0\), \(z=c-a=x+y\). Then \(xyz=2\). By AM-GM, \(4xy\le(x+y)^2=z^2\), hence \(2=xyz\le\frac{z^3}{4}\), so \(z\ge2\). If \(z=2\), equality in \(4xy\le z^2\) gives \(x=y=1\). Then \(b=a+1>1\), \(c=a+2>2\), and \(a^2b+b^2c+c^2a>b^2c>2\), contradicting the first equality. Therefore \(z>2\), i.e. \(c-a>2\).

A. Source analysis. Main objects: positive quantities, a product or a sum, choosing the right terms for AM-GM, and a strict equality case. The obvious first attempt usually applies AM-GM to the visible terms, but it misses the hidden structure. The hidden observation is to normalize, order, multiply, or substitute first so that the product of the chosen terms is controlled.

B. New problem. The statement is rewritten; only the method architecture is preserved.

C-D. Hints. The hint fields contain a gentle hint and a method-level hint.

E. Full solution. Given in the solution field.

F. Difficulty justification. Regional level 7: it requires finding a hidden product of differences and handling equality.

G. Why this is not a one-step exercise. Even after factorization, AM-GM and a separate equality exclusion are needed.