Dan Sitaru's Cyclic Inequality in Three Variables IV

Problem

Prove that if $x,y,z\gt 0$ and $x+y+z=3\,$ then

$\displaystyle \sum_{cycl}\frac{(x+y)z}{\sqrt{4x^2+xy+4y^2}}\le 2.$

Solution

$\displaystyle \begin{align} 4x^2+xy+4y^2 &= 2[(x+y)2+(x-y)^2]+\frac{1}{4}[(x+y)^2-(x-y)2)]\\ &=\frac{9}{4}(x+y)^2+\frac{7}{4}(x-y)^2\ge\frac{9}{4}(x+y)^2\\ &\Rightarrow\;\sqrt{4x^2+xy+4y^2}\ge \frac{3}{2}(x+y)\\ &\Rightarrow\;\frac{(x+y)z}{\sqrt{4x^2+xy+4y^2}}\le \frac{2}{3}z\\ &\Rightarrow\;\sum_{cycl}\frac{(x+y)z}{\sqrt{4x^2+xy+4y^2}}\le \frac{2}{3}\sum_{cycl}z=2. \end{align}$

Equality is attained for $x=y=z=1.$

Acknowledgment

Dan Sitaru has kindly posted problem of his at the CutTheKnotMath facebook page, with a solution by Ravi Prakash. The problem appeared earlier at the Romanian Mathematical Magazine.

 

Inequalities with the Sum of Variables as a Constraint

|Contact| |Up| |Front page| |Contents| |Algebra| |Store|

Copyright © 1996-2017 Alexander Bogomolny

 62015652

Search by google: