Problem
ALG-B2-M03-P019 Quadratic substitution
#19
★★★★★ Level 5 of 5
Let \(x,y,z>0\). Prove \[\frac{x^4}{x^2+xy+y^2}+\frac{y^4}{y^2+yz+z^2}+\frac{z^4}{z^2+zx+x^2}\ge\frac{x^2+y^2+z^2}{3}.\]
Hint 1. Apply Cauchy to numerators \(x^2,y^2,z^2\).
Hint 2. Compare the denominator sum with \(3(x^2+y^2+z^2)\).
By Cauchy, the left side is at least \[\frac{(x^2+y^2+z^2)^2}{2(x^2+y^2+z^2)+xy+yz+zx}.\] Since \(xy+yz+zx\le x^2+y^2+z^2\), the denominator is at most \(3(x^2+y^2+z^2)\). This gives the required bound \(\frac{x^2+y^2+z^2}{3}\).
Cauchy-Schwarz module training problem. Method tags: cauchy, substitution.