Let $ x$, $ y$, $ z$ be positive numbers. Find the minimum value of: $ (a)\quad \frac{x^2 + y^2 + z^2}{xy + yz}$ $ (b)\quad \frac{x^2 + y^2 + 2z^2}{xy + yz}$
Problem
Source: Croatian Team Selection Test 2008
Tags: inequalities proposed, inequalities
dduclam
22.04.2008 09:14
April wrote: Let $ x$, $ y$, $ z$ be positive numbers. Find the minimum value of: $ (a)\quad \frac {x^2 + y^2 + z^2}{xy + yz}$ $ (b)\quad \frac {x^2 + y^2 + 2z^2}{xy + yz}$
By AM-GM:
(a) $ x^2 + y^2 + z^2=x^2+\frac{y^2}2+\frac{y^2}2+z^2\ge\sqrt2 (xy+yz)$
(b) $ x^2 + y^2 + 2z^2=x^2+\frac{2y^2}3+\frac{y^2}3+2z^2\ge2\sqrt{\frac{2}3}(xy+yz)$
nsato
22.04.2008 10:31
Looks familiar... http://www.artofproblemsolving.com/Forum/viewtopic.php?t=200220
jasperE3
11.06.2022 03:10
$(\text a)$ We have, by AM-GM: $$x^2+y^2+z^2=x^2+\frac12y^2+\frac12y^2+z^2\ge xy\sqrt 2+yz\sqrt 2$$so $\frac{x^2+y^2+z^2}{xy+yz}\ge\sqrt2$, with equality when $(x,y,z)=\left(\sqrt2,2,\sqrt2\right)$. $(\text b)$ We have, by AM-GM: $$x^2+y^2+2z^2=x^2+\frac23y^2+\frac13y^2+2z^2\ge\frac{2\sqrt6}3xy+\frac{2\sqrt6}3yz$$so $\frac{x^2+y^2+2z^2}{xy+yz}\ge\frac{2\sqrt6}3$, with equality when $(x,y,z)=\left(2\sqrt6,6,\sqrt6\right)$.