Problem
ALG-B2-M03-P018 Cyclic fourth powers
Let \(a,b,c>0\), \(a+b+c=3\). Prove \[\frac{a}{b^4+2b}+\frac{b}{c^4+2c}+\frac{c}{a^4+2a}\ge1.\]
Hint 1. Transform \(\frac{a}{b^4+2b}\) using \(\frac{a}{b}\).
Hint 2. Use \(b^3+2\ge3b\), then \(ab+bc+ca\le3\).
\[\frac{a}{b^4+2b}=\frac{a}{b}-\frac{2ab^2}{b^3+2}\ge\frac{a}{b}-\frac{2ab^2}{3b}=\frac{a}{b}-\frac23ab,\] because \(b^3+2=b^3+1+1\ge3b\). Adding cyclically, \[\sum\frac{a}{b^4+2b}\ge\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)-\frac23(ab+bc+ca).\] The first bracket is at least \(3\) by AM-GM. Also, \(a+b+c=3\) implies \(ab+bc+ca\le3\). Thus the right side is at least \(3-\frac23\cdot3=1\).
A. Source analysis. Main objects: fractional sums, cyclic denominators, Engel form, or a preliminary denominator estimate. The obvious move is to estimate each fraction separately. The hidden observation is to first obtain the right denominator sum or transform the fraction so that Cauchy gives global control.
B. New problem. The statement is rewritten as a training problem while preserving the solution architecture.
C-D. Hints. The hint fields contain a gentle and a method-level hint.
E. Full solution. See solution.
F. Difficulty justification. Regional level 7: it requires transforming the fraction, applying AM-GM, and then using a Cauchy/square estimate for \(ab+bc+ca\).
G. Why this is not a one-step exercise. None of the three estimates solves the problem alone; they must be chained together.