Problem
ALG-B2-M02-P017 Product of differences
Real \(x,y,z\) satisfy \(x^2+y^2+z^2=1\). Prove \[(x-y)(y-z)(x-z)\le\frac1{\sqrt2}.\]
Hint 1. If the left side is negative, it is clear; otherwise order \(x\ge y\ge z\).
Hint 2. Set \(u=x-y\), \(v=y-z\).
We may assume \(x\ge y\ge z\). Let \(u=x-y\ge0\), \(v=y-z\ge0\), so \(x-z=u+v\), and the left side is \(uv(u+v)\). From \(x^2+y^2+z^2=1\), we get \((x-y)^2+(y-z)^2+(x-z)^2\le3\), hence \(u^2+v^2+(u+v)^2\le3\). For fixed \(s=u+v\), \(uv\le\frac{s^2}{4}\). Also \(u^2+v^2\ge\frac{s^2}{2}\), so \(\frac32s^2\le3\), i.e. \(s\le\sqrt2\). Therefore \(uv(u+v)\le\frac{s^3}{4}\le\frac{(\sqrt2)^3}{4}=\frac1{\sqrt2}\).
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: it requires ordering variables and optimizing through two differences.
G. Why this is not a one-step exercise. AM-GM is applied not to the original variables, but to differences after substitution.