Problem
ALG-B2-M02-P018 Cyclic sum of fractions
Let \(x_1,\ldots,x_n>0\), \(n\ge2\), and \(x_{n+1}=x_1\). Prove \[\sum_{i=1}^{n}\frac{1+x_i^2}{1+x_ix_{i+1}}\ge n.\]
Hint 1. First prove an estimate for the product of all fractions.
Hint 2. Use \((1+x_i^2)(1+x_{i+1}^2)\ge(1+x_ix_{i+1})^2\).
For each \(i\), \[(1+x_i^2)(1+x_{i+1}^2)-(1+x_ix_{i+1})^2=(x_i-x_{i+1})^2\ge0.\] Multiplying over all \(i\), we obtain \[\prod_{i=1}^{n}\frac{1+x_i^2}{1+x_ix_{i+1}}\ge1.\] Applying AM-GM to the \(n\) positive fractions, their sum is at least \(n\) times the \(n\)-th root of the product, hence at least \(n\).
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: the key step is to obtain a product estimate first, not estimate the sum directly.
G. Why this is not a one-step exercise. The visible sum is not ready for AM-GM until the product is controlled.