Problem
ALG-B2-M02-P019 Three radicals with denominators
Let \(a,b,c\ge1\). Prove \[\frac{a+b+c}{4}\ge\frac{\sqrt{ab-1}}{b+c}+\frac{\sqrt{bc-1}}{c+a}+\frac{\sqrt{ca-1}}{a+b}.\]
Hint 1. Estimate the first term using \(b+c\ge2\sqrt{bc}\).
Hint 2. Then apply AM-GM again to \(2\sqrt{\left(a-\frac1b\right)\frac1c}\).
We have \(b+c\ge2\sqrt{bc}\), hence \[4\frac{\sqrt{ab-1}}{b+c}\le2\sqrt{\frac{ab-1}{bc}}=2\sqrt{\left(a-\frac1b\right)\frac1c}\le a-\frac1b+\frac1c.\] Similarly, \[4\frac{\sqrt{bc-1}}{c+a}\le b-\frac1c+\frac1a,\qquad 4\frac{\sqrt{ca-1}}{a+b}\le c-\frac1a+\frac1b.\] Adding gives that four times the right side is at most \(a+b+c\), which is the desired inequality.
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: AM-GM is applied twice in each cyclic term.
G. Why this is not a one-step exercise. The first AM-GM only handles the denominator; the second turns the radical into a telescoping sum.