Problem
ALG-B2-M02-P025 Three shifted factors
Numbers \(a,b,c>1\) satisfy \[\left(a-\frac1b\right)\left(b-\frac1c\right)\left(c-\frac1a\right)=1.\] Prove \[\left(a-\frac1a\right)^2+\left(b-\frac1b\right)^2+\left(c-\frac1c\right)^2\ge\frac ba+\frac cb+\frac ac.\]
Hint 1. Multiply the three factors by \(\frac ba\), \(\frac cb\), \(\frac ac\).
Hint 2. Obtain a new product equal to \(1\), then apply AM-GM to squares.
Multiplying the three factors by \(\frac ba,\frac cb,\frac ac\), we get \[\left(b-\frac1a\right)\left(c-\frac1b\right)\left(a-\frac1c\right)=1.\] All three factors are positive. By AM-GM, \[\left(b-\frac1a\right)^2+\left(c-\frac1b\right)^2+\left(a-\frac1c\right)^2\ge3.\] Expanding gives \[a^2+b^2+c^2+\frac1{a^2}+\frac1{b^2}+\frac1{c^2}\ge3+2\left(\frac ba+\frac cb+\frac ac\right).\] Also, \(\frac ba+\frac cb+\frac ac\ge3\). Hence the right side is at least \(6+\frac ba+\frac cb+\frac ac\). Expanding the left side of the desired inequality gives \(a^2+b^2+c^2+\frac1{a^2}+\frac1{b^2}+\frac1{c^2}-6\). The result follows.
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: two AM-GM applications and a preliminary cyclic multiplication are needed.
G. Why this is not a one-step exercise. The original product is not the one needed for the squares; it must be cyclically transformed first.