A Cyclic Inequality in Three Variables XIV

Problem

A Cyclic Inequality  in  Three Variables  XIV

Solution

With the Cauchy-Schwarz inequality, we get

$\displaystyle\begin{align} \sum_{cycl}\frac{xy}{xy+y^2+zx} &= \sum_{cycl}\frac{\displaystyle xy\left(\frac{x}{y}+1+\frac{z}{x}\right)}{\displaystyle (xy+y^2+zx)\left(\frac{x}{y}+1+\frac{z}{x}\right)}\\ &\le\sum_{cycl}\frac{x^2+xy+yz}{(x+y+z)^2}\\ &=\frac{(x+y+z)^2}{(x+y+z)^2}\\ &=1. \end{align}$

Acknowledgment

This problem has been shared by Mihai Dicu at the Olimpiada pe Scoala (The School Yard Olympiad) facebook group. The beautiful solution is by Diego Alvariz.

 

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