Problem
ALG-B2-M01-P025 Chain of signs
Numbers \(a,b,c\) satisfy \(a^2+b^2<(a-b)^2\) and \(b^2+c^2<(b-c)^2\). Prove that \(a^4+c^4<(a+c)^4\).
Hint 1. Expand the first two inequalities.
Hint 2. First determine the signs of \(ab\), \(bc\), then \(ac\).
From \(a^2+b^2<(a-b)^2=a^2-2ab+b^2\), we get \(ab<0\). Similarly, \(bc<0\). Thus \(a\) and \(b\) have opposite signs, and \(b\) and \(c\) have opposite signs; hence \(a\) and \(c\) have the same sign, so \(ac>0\). Then \[(a+c)^4-a^4-c^4=4a^3c+6a^2c^2+4ac^3=2ac(2a^2+3ac+2c^2)>0,\] because \(ac>0\) and \(2a^2+3ac+2c^2>0\). Therefore \(a^4+c^4<(a+c)^4\).
A. Source analysis. Main objects: inequalities, order, an extremal element, or an invariant. The obvious first move usually gives only a local estimate. The hidden observation is to choose the right nondecreasing quantity, or to add/multiply inequalities only after signs are controlled. The needed step is an ordering, an invariant, a product transformation, or a boundary case.
F. Difficulty justification. Regional level 6: the problem is short, but the hidden object is the signs of products.
G. Why this is not a one-step exercise. Expanding only starts the solution; the signs must be chained together.