Problem
ALG-B2-M02-P012 Sum and product of four numbers
Positive \(a,b,c,d\) satisfy \(2(a+b+c+d)\ge abcd\). Prove \[a^2+b^2+c^2+d^2\ge abcd.\]
Hint 1. Split into \(abcd\ge16\) and \(abcd<16\).
Hint 2. In the first case use the condition; in the second use AM-GM for the squares.
Let \(P=abcd\), \(S=a+b+c+d\). If \(P\ge16\), then \[a^2+b^2+c^2+d^2\ge4\left(\frac{S}{4}\right)^2\ge4\left(\frac{P}{8}\right)^2=\frac{P^2}{16}\ge P.\] If \(P<16\), then by AM-GM \[a^2+b^2+c^2+d^2\ge4\sqrt[4]{a^2b^2c^2d^2}=4\sqrt P.\] Since \(P<16\), \(4\sqrt P>P\). The result follows in both cases.
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. Final level 8: the solution requires choosing the threshold \(16\) and switching between two AM-GM applications.
G. Why this is not a one-step exercise. Direct AM-GM does not use the condition; the condition is useful only in one case.