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CaptainBlack · #10 June 25th, 2007, 10:06 PM

Quote:

Originally Posted by mathisfun1

$x^2+y^2+z^2 \ge xy + yz + zx$ results from AM-GM inequality.
Inequality of arithmetic and geometric means - Wikipedia, the free encyclopedia

Show us how.

I see how it follows from the Cauchy Scwartz inequality:

$| \bold{x} \cdot \bold{y} |\le \| \bold{x} \|\ \| \bold{y} \|$

Then putting $\bold{x}=(a,b,c)$ and $\bold{y}=(b,c,a)$ , with $a, b, c \in \mathbb{R}$ , we have:

$ab + bc + ca \le |ab + bc + ca| \le \sqrt{a^2+b^2+c^2}\ \sqrt{b^2+c^2+a^2} = a^2+b^2+c^2$

RonL