document.write( "Question 27846: If x, y, and z are reall numbers, use the Cauchy-Schwarz inequality to show that (x+y+z)^2/3<=x^2+y^2+z^2. \n" ); document.write( "

Algebra.Com's Answer #15196 by venugopalramana(3286) $\"\"$ $\"About$

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(x+y+z)^2/3<=x^2+y^2+z^2.
\n" ); document.write( "AS PER Cauchy-Schwarz inequality
\n" ); document.write( " IF A AND B ARE 2 VECTORS THEN
\n" ); document.write( "|A.B| <=|A|.|B|
\n" ); document.write( "LET US TAKE A AND B AS 3 DIMENSIONAL VECTORS IN A VECTOR SPACE AS
\n" ); document.write( "A=XI+YJ+ZK...AND
\n" ); document.write( "B=I+J+K
\n" ); document.write( "WHERE I,J AND K ARE UNIT VECTORS ALONG THE 3 MUTUALLY PERPENDICULAR DIRECTIONS.HENCE WE GET FROM THE ABOVE
\n" ); document.write( "|(XI+YJ+JK).(I+J+K)|<= |(XI+YJ+JK)|.|(I+J+K)|
\n" ); document.write( "X+Y+Z<={(X^2+Y^2+Z^2)^0.5}.{(1^2+1^2+1^2)^0.5}={3(X^2+Y^2+Z^2)}^0.5
\n" ); document.write( "SQUARING BOTH SIDES
\n" ); document.write( "(X+Y+Z)^2<=3(X^2+Y^2+Z^2)
\n" ); document.write( "{(X+Y+Z)^2}/3<=(X^2+Y^2+Z^2)
\n" ); document.write( " \n" ); document.write( "

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