Thread: Problem 28
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Old June 25th, 2007, 11:36 PM
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The inequality x^2+y^2+z^2\geq xy+yz+zx can be proved like this:
Multiplying with 2, the inequality is equivalent to
2x^2+2y^2+2z^2\geq 2xy+2yz+2zx\Leftrightarrow
\Leftrightarrow (x^2-2xy+y^2)+(y^2-2yz+z^2)+(z^2-2zx+x^2)\geq 0\Leftrightarrow
\Leftrightarrow (x-y)^2+(y-z)^2+(z-x)^2\geq 0.