Find smallest value of (x^2 + y^2 + z^2)/(xyz)

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Find smallest value of (x^2 + y^2 + z^2)/(xyz)

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Source: CWMO 2001, Problem 4

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3647 posts

orl

#1 h Dec 27, 2008, 1:29 AM • 2 Y

Y by Adventure10, Mango247

Let $x, y, z$ be real numbers such that $x + y + z \geq xyz$ . Find the smallest possible value of $\frac {x^2 + y^2 + z^2}{xyz}$ .

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#2 h Dec 27, 2008, 3:39 AM • 3 Y

Y by vsathiam, Adventure10, Mango247

$x=y=1$ and $z<0$ will give $-\infty$ when $z$ is close to $0$ .

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Erken

#3 h Dec 27, 2008, 5:33 AM • 1 Y

Y by Adventure10

okay, but what if $x,y,z$ are positive reals?

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arqady

#4 h Dec 27, 2008, 2:27 PM • 2 Y

Y by Adventure10, Mango247

orl wrote:

Let $x, y, z$ be real numbers such that $x + y + z \geq xyz$ . Find the smallest possible value of $\frac {x^2 + y^2 + z^2}{xyz}$ .

See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=17142

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Dr Sonnhard Graubner

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Dr Sonnhard Graubner

#5 h Dec 27, 2008, 3:02 PM • 2 Y

Y by Adventure10, Mango247

hello, if the variables are positive the searched minimum is $\sqrt{3}$ .
Sonnhard.

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Rofler

#6 h Dec 27, 2008, 8:25 PM • 4 Y

Y by vsathiam, Adventure10, Mango247, and 1 other user

Let $x + y + z = kxyz$

Then

$\sqrt {(\frac {x^2 + y^2 + z^2}{xyz})^2} = \sqrt {\frac {x^4 + y^4 + z^4 + 2x^2y^2 + 2y^2z^2 + 2z^2x^2}{x^2y^2z^2}} = \sqrt {\frac {x^4 + y^4 + z^4 + 2x^2y^2 + 2y^2z^2 + 2z^2x^2}{(xyz)(\frac {x + y + z}{k})}}$

Hence, we need to minimize

$\frac {x^4 + y^4 + z^4 + 2x^2y^2 + 2y^2z^2 + 2z^2x^2}{(xyz)(\frac {x + y + z}{k})}$

I claim that $x^4 + y^4 + z^4 + 2x^2y^2 + 2y^2z^2 + 2z^2x^2 \ge 3xyz(x + y + z)$

Which is true by 2 trivial applications of muirheads inequality.

Hence, since $3xyz(x + y + z) \ge 3xyz(\frac {x + y + z}{k})$ , we have:

$\frac {x^4 + y^4 + z^4 + 2x^2y^2 + 2y^2z^2 + 2z^2x^2}{(xyz)(\frac {x + y + z}{k})}\ge 3$

equality will hold when $x = y = z = \sqrt {3}$ . The final minimum is the square root of this answer, which is $\sqrt {3}$ .

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WakeUp

#7 h Jan 26, 2011, 10:24 PM • 2 Y

Y by Adventure10, Mango247

Actually to show that $x^2+y^2+z^2\ge\sqrt{3}xyz$ 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).

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