Problem
ALG-B2-M02-P015 Fractional linear substitution
Real numbers \(u_1,u_2,u_3,u_4\) have absolute value greater than \(1\), and \[\prod_{i=1}^{4}\frac{u_i+1}{u_i-1}=1.\] Prove \[\frac1{u_1-1}+\frac1{u_2-1}+\frac1{u_3-1}+\frac1{u_4-1}>0.\]
Hint 1. Set \(x_i=\frac{u_i+1}{u_i-1}\).
Hint 2. Then \(x_1x_2x_3x_4=1\), and \(\frac1{u_i-1}=\frac{x_i-1}{2}\).
Since \(|u_i|>1\), the numbers \(x_i=\frac{u_i+1}{u_i-1}\) are positive. Their product is \(1\). Also, \(\frac1{u_i-1}=\frac{x_i-1}{2}\). Thus it is enough to prove \(x_1+x_2+x_3+x_4>4\). By AM-GM, \(x_1+x_2+x_3+x_4\ge4\sqrt[4]{x_1x_2x_3x_4}=4\). Equality would require all \(x_i=1\), impossible for finite \(u_i\). Hence the sum is strictly greater than \(4\), and the required sum is positive.
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. Final level 8: it needs a non-obvious fractional linear substitution and a strict equality case.
G. Why this is not a one-step exercise. AM-GM becomes applicable only after turning the condition into a product of positive new variables.