SOLUTION: Show that (3x+2y+z)^2<= 3(9x^2+4y^2+z^2) where x, y, and z are real numbers

Question 27847: Show that (3x+2y+z)^2<= 3(9x^2+4y^2+z^2) where x, y, and z are real numbers
Answer by venugopalramana(3286) (Show Source): You can put this solution on YOUR website!
Show that (3x+2y+z)^2<= 3(9x^2+4y^2+z^2) where x, y, and z are real numbers
LET US FIRST PROVE THIS INEQUALITY.....
(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)
IF WE PUT IN THIS INEQUALITY
X=3x...;Y=2y.....;Z=z...WE GET ...
{(3x+2y+z)^2}/3<={(3x)^2+(2y)^2+z^2}=(9x^2+4y^2+z^2)

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