Problem
ALG-B2-M02-P024 Three strict inequalities
Real numbers \(x,y,z\) satisfy \[2x>y^2+z^2,\qquad 2y>z^2+x^2,\qquad 2z>x^2+y^2.\] Prove that \(x<1\), \(y<1\), \(z<1\).
Hint 1. Add two inequalities to bound the third variable.
Hint 2. After rearrangement, squares of the form \((x-1)^2\) appear.
Add the first two inequalities: \(2x+2y>x^2+y^2+2z^2\). Rearranging, \[0>(x-1)^2+(y-1)^2+2(z^2-1).\] Since the first two terms are nonnegative, \(z^2<1\), hence \(z<1\). Similarly, adding the other pairs gives \(x<1\) and \(y<1\).
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: one must choose the right pairs of conditions and complete squares.
G. Why this is not a one-step exercise. No single inequality bounds its own variable above by \(1\).