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)![]() ![]() You can put this solution on YOUR website! Consider the vectors (x, y, z) and ( \n" ); document.write( "Then direct application of the Cauchy-Schwartz Inequality gives:\r \n" ); document.write( "\n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "we don't even need the non-negativity conditions for x, y, and z. \n" ); document.write( " |