Problem
ALG-B2-M02-P014 Roots and the condition \(ab+bc+ca\)
Positive \(a,b,c\) satisfy \(ab+bc+ca=2\). Prove \[\sqrt{a+\frac2a}+\sqrt{b+\frac2b}+\sqrt{c+\frac2c}\ge2(\sqrt a+\sqrt b+\sqrt c).\]
Hint 1. Prove separately \(\sqrt{a+\frac2a}\ge\sqrt b+\sqrt c\).
Hint 2. Replace \(2\) by \(ab+bc+ca\).
Since \(ab+bc+ca=2\), \[a+\frac2a=a+\frac{ab+ac+bc}{a}=a+b+c+\frac{bc}{a}.\] By AM-GM, \(a+\frac{bc}{a}\ge2\sqrt{bc}\). Hence \(a+\frac2a\ge b+c+2\sqrt{bc}=(\sqrt b+\sqrt c)^2\), so \(\sqrt{a+\frac2a}\ge\sqrt b+\sqrt c\). Adding the three cyclic estimates gives the result.
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: the key step is inserting the condition into each radical.
G. Why this is not a one-step exercise. Applying AM-GM to the original sum directly does not create the needed pairs under the radicals.