Problem
ALG-B2-M01-P011 Difference of powers
Numbers \(a,b\) satisfy \(a^3-b^3=3\) and \(a^5-b^5\ge9\). Prove that \(a^2+b^2\ge3\).
Hint 1. Multiply \(a^2+b^2\) by \(a^3-b^3\).
Hint 2. After expansion, a nonnegative remainder appears.
From \(a^3-b^3=3\) we get \(a>b\). Then \[3(a^2+b^2)=(a^2+b^2)(a^3-b^3)=a^5-b^5+a^2b^2(a-b).\] The right side is at least \(9\), because \(a^5-b^5\ge9\) and \(a^2b^2(a-b)\ge0\). Hence \(3(a^2+b^2)\ge9\), so \(a^2+b^2\ge3\).
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 key step is guessing the right product of powers.
G. Why this is not a one-step exercise. The conclusion does not follow from direct power comparison; an exact algebraic remainder is needed.