# Cyclic Inequality In Three Variables From Vietnam by Rearrangement

### Solution

Assuming, WLOG, $x\ge y\ge z,$

$\displaystyle x^2+y^2\ge x^2+z^2\ge y^2+z^2\\ x^3+y^3\ge x^3+z^3\ge y^3+z^3,$

implying, by the rearrangement theorem,

$\displaystyle \sum_{cycl}\frac{x^3+y^3}{y^2+z^2}\ge \sum_{cycl}\frac{x^3+y^3}{x^2+y^2}.$

On the other hand, $\displaystyle x^3+y^3\ge\frac{(x+y)(x^2+y^2)}{2},$ so that

$\displaystyle \sum_{cycl}\frac{x^3+y^3}{x^2+y^2}\ge x+y+z.$

But, by rearrangement after squaring,

$x+y+z\ge\sqrt{3(xy+yz+zx)}=3.$

### Acknowledgment

This problem by Nguyen Viet Hung was kindly communicated to me by Leo Giugiuc, along with a solution of his. The problem was originally posted at the Olimpiada pe Scoala (The School Yard Olympiad) facebook group.