Let x,y,z be real numbers such that x+y+z≥xyz. Find the smallest possible value of x2+y2+z2xyz.
Problem
Source: CWMO 2001, Problem 4
Tags: inequalities, inequalities unsolved
27.12.2008 05:39
x=y=1 and z<0 will give −∞ when z is close to 0.
27.12.2008 07:33
okay, but what if x,y,z are positive reals?
27.12.2008 16:27
orl wrote: Let x,y,z be real numbers such that x+y+z≥xyz. Find the smallest possible value of x2+y2+z2xyz. See here: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=17142
27.12.2008 17:02
hello, if the variables are positive the searched minimum is √3. Sonnhard.
27.12.2008 22:25
Let x+y+z=kxyz Then √(x2+y2+z2xyz)2=√x4+y4+z4+2x2y2+2y2z2+2z2x2x2y2z2=√x4+y4+z4+2x2y2+2y2z2+2z2x2(xyz)(x+y+zk) Hence, we need to minimize x4+y4+z4+2x2y2+2y2z2+2z2x2(xyz)(x+y+zk) I claim that x4+y4+z4+2x2y2+2y2z2+2z2x2≥3xyz(x+y+z) Which is true by 2 trivial applications of muirheads inequality. Hence, since 3xyz(x+y+z)≥3xyz(x+y+zk), we have: x4+y4+z4+2x2y2+2y2z2+2z2x2(xyz)(x+y+zk)≥3 equality will hold when x=y=z=√3. The final minimum is the square root of this answer, which is √3.
27.01.2011 00:24
Actually to show that x2+y2+z2≥√3xyz appeared on the Balkan 2001 paper http://www.artofproblemsolving.com/Forum/resources.php?c=1&cid=19&year=2001&sid=5c22a3c467b0fb6290e11cdb1f19768c (this problem is Chinese Olympiad 2001).