SOLUTION: 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.

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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.
Answer by venugopalramana(3286) About Me  (Show Source):
You can put this solution on YOUR website!
(x+y+z)^2/3<=x^2+y^2+z^2.
AS PER Cauchy-Schwarz inequality
IF A AND B ARE 2 VECTORS THEN
|A.B| <=|A|.|B|
LET US TAKE A AND B AS 3 DIMENSIONAL VECTORS IN A VECTOR SPACE AS
A=XI+YJ+ZK...AND
B=I+J+K
WHERE I,J AND K ARE UNIT VECTORS ALONG THE 3 MUTUALLY PERPENDICULAR DIRECTIONS.HENCE WE GET FROM THE ABOVE
|(XI+YJ+JK).(I+J+K)|<= |(XI+YJ+JK)|.|(I+J+K)|
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
SQUARING BOTH SIDES
(X+Y+Z)^2<=3(X^2+Y^2+Z^2)
{(X+Y+Z)^2}/3<=(X^2+Y^2+Z^2)