document.write( "Question 461593: prove that for x,y,z>=0, x^2+y^2+z^2>=xy+yz+zx \n" ); document.write( "
Algebra.Com's Answer #316526 by robertb(5830)\"\" \"About 
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Consider the vectors (x, y, z) and ( \"1%2Fsqrt%283%29\", \"1%2Fsqrt%283%29\",\"1%2Fsqrt%283%29\").
\n" ); document.write( "Then direct application of the Cauchy-Schwartz Inequality gives:\r
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\n" ); document.write( "\n" ); document.write( "\"%28x%5E2+%2B+y%5E2+%2B+z%5E2%29%2A1+%3E=+%281%2F3%29%28x%2By%2Bz%29%5E2\"\r
\n" ); document.write( "\n" ); document.write( "\"3x%5E2+%2B+3y%5E2+%2B+3z%5E2+%3E=+x%5E2+%2B+y%5E2+%2B+z%5E2++%2B+2xy+%2B+2xz+%2B+2yz\"\r
\n" ); document.write( "\n" ); document.write( "\"2x%5E2+%2B2y%5E2+%2B+2z%5E2+%3E=++2xy+%2B+2xz+%2B+2yz\"
\n" ); document.write( "\"x%5E2+%2By%5E2+%2B+z%5E2+%3E=++xy+%2B+xz+%2B+yz\".\r
\n" ); document.write( "\n" ); document.write( "we don't even need the non-negativity conditions for x, y, and z.
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