SOLUTION: show that x^2y+y^2z+z^2x>=3xyz for positive real numbers x,y, and z.

Question 27845: show that x^2y+y^2z+z^2x>=3xyz for positive real numbers x,y, and z.
Answer by venugopalramana(3286) (Show Source): You can put this solution on YOUR website!
TST x^2y+y^2z+z^2x>=3xyz
WE KNOW THAT ARITHMATIC MEAN >=GEOMETRIC MEAN FOR POSITIVE NUMBERS.
CONSIDER X^2Y,Y^Z,Z^X ...3 NUMBERS..THEY ARE ALL POSITIVE SINCE X,Y,Z ARE POSITIVE.
SO
(x^2y+y^2z+z^2x)/3>=(x^2y*y^2z*z^2x)^(1/3)=(X^3*Y^3*Z^3)^(1/3)=XYZ
HENCE
x^2y+y^2z+z^2x>=3xyz

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