Problem
ALG-B2-M02-P016 Four reciprocal squares
Positive \(a,b,c,d\) have sum \(3\). Prove \[\frac1{a^2}+\frac1{b^2}+\frac1{c^2}+\frac1{d^2}\le\frac1{a^2b^2c^2d^2}.\]
Hint 1. Multiply by \(a^2b^2c^2d^2\).
Hint 2. After ordering, use \(ab(c+d)\le1\).
After multiplying, it is enough to prove \[a^2b^2c^2+a^2b^2d^2+a^2c^2d^2+b^2c^2d^2\le1.\] Assume \(a\ge b\ge c\ge d\). By AM-GM for \(a,b,c+d\), \(ab(c+d)\le\left(\frac{a+b+c+d}{3}\right)^3=1\), so \(a^2b^2(c+d)^2\le1\). It remains to check that the left side is at most \(a^2b^2(c+d)^2\). Expanding, this reduces to \(c^2d^2(a^2+b^2)\le2a^2b^2cd\), which follows from \(cd\le b^2\) and \(cd\le a^2\). The original inequality follows.
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: it needs multiplication, ordering, and a non-obvious choice of three AM-GM terms.
G. Why this is not a one-step exercise. Estimating fractions separately does not work; one must pass to a symmetric polynomial.