Problem
ALG-B2-M02-P013 Four squares
Real \(a,b,c,d\) satisfy \(a^2+b^2+c^2+d^2=9\). Prove \((3+a)(3+b)\ge cd\).
Hint 1. First estimate \(cd\) through \(c^2+d^2\).
Hint 2. A square \((a+b+3)^2\) will remain.
We have \(cd\le |cd|\le\frac{c^2+d^2}{2}=\frac{9-a^2-b^2}{2}\). It is enough to prove \(2(3+a)(3+b)\ge9-a^2-b^2\). After moving all terms to the left, we get \(a^2+2ab+b^2+6a+6b+9=(a+b+3)^2\ge0\). Hence the inequality is true.
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 6: one AM-GM estimate and a hidden square after rearrangement.
G. Why this is not a one-step exercise. One must first identify which part should be replaced by a sum of squares.