Problem
ALG-B2-M02-P022 Twelve squares from one square
Let \(a_1\ge a_2\ge\cdots\ge a_{18}>0\) and \(a_1^2+\cdots+a_{18}^2=1\). Prove \[a_7+a_{10}+a_{13}+a_{16}\le1,\qquad a_7+a_8+a_9\le1.\]
Hint 1. Use the ordering: the first \(16\) squares can be grouped in fours.
Hint 2. The last step is RMS-AM.
Since the sequence is nonincreasing, \[a_1^2+\cdots+a_{16}^2\ge4a_7^2+4a_{10}^2+4a_{13}^2+4a_{16}^2.\] Hence \[1\ge4(a_7^2+a_{10}^2+a_{13}^2+a_{16}^2)\ge(a_7+a_{10}+a_{13}+a_{16})^2,\] by RMS-AM in the last step. Thus the first sum is at most \(1\). Similarly, \(1\ge3(a_7^2+a_8^2+a_9^2)\ge(a_7+a_8+a_9)^2\), which gives the second inequality.
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 problem isolates the algebraic core of a packing argument through ordering and RMS-AM.
G. Why this is not a one-step exercise. Plain AM-GM over all \(18\) numbers does not see the needed indices; the ordered list must be grouped correctly.