Problem
ALG-B2-M03-P017 Four cubic fractions
Let \(a,b,c,d>0\) and \(a+b+c+d=8\). Prove \[\frac{a^3}{a^2+b+c}+\frac{b^3}{b^2+c+d}+\frac{c^3}{c^2+d+a}+\frac{d^3}{d^2+a+b}\ge4.\]
Hint 1. First prove \(\frac{a^3}{a^2+b+c}\ge a-\frac{\sqrt{b+c}}2\).
Hint 2. Then estimate the sum of the four roots using RMS-AM.
As in the previous problem, \(\frac{a^3}{a^2+b+c}\ge a-\frac{\sqrt{b+c}}2\), and similarly for the other terms. Thus the left side is at least \[8-\frac12(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+d}+\sqrt{d+a}).\] It remains to show that the root sum is at most \(8\). By Cauchy, \[(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+d}+\sqrt{d+a})^2\le4\cdot2(a+b+c+d)=64.\] Hence the root sum is at most \(8\), and the left side is at least \(4\).
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: the solution combines a preliminary denominator estimate with Cauchy/RMS-AM for roots.
G. Why this is not a one-step exercise. Direct Cauchy on the original sum does not immediately give the constant; an intermediate linear estimate is required.