Problem
ALG-B2-M01-P017 Fifth and third powers
Let \(x,y>0\) and \(x^5-y^3\ge4x\). Prove that \(x^3\ge\sqrt[3]{16}\,y\).
Hint 1. Cube the desired inequality.
Hint 2. It is enough to prove \(x^9\ge16x^5-64x\).
We need to prove \(x^9\ge16y^3\). From the condition, \(y^3\le x^5-4x\), so it is enough to show \(x^9\ge16x^5-64x\). But \[x^9-16x^5+64x=x(x^4-8)^2\ge0,\] since \(x>0\). Therefore \(x^9\ge16y^3\), hence \(x^3\ge\sqrt[3]{16}\,y\).
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 7: one must choose the right power and see a square after rearrangement.
G. Why this is not a one-step exercise. Direct monotonicity of powers does not produce the required constant.