document.write( "Question 27845: show that x^2y+y^2z+z^2x>=3xyz for positive real numbers x,y, and z. \n" ); document.write( "
Algebra.Com's Answer #15192 by venugopalramana(3286)\"\" \"About 
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TST x^2y+y^2z+z^2x>=3xyz
\n" ); document.write( "WE KNOW THAT ARITHMATIC MEAN >=GEOMETRIC MEAN FOR POSITIVE NUMBERS.
\n" ); document.write( "CONSIDER X^2Y,Y^Z,Z^X ...3 NUMBERS..THEY ARE ALL POSITIVE SINCE X,Y,Z ARE POSITIVE.
\n" ); document.write( "SO
\n" ); document.write( "(x^2y+y^2z+z^2x)/3>=(x^2y*y^2z*z^2x)^(1/3)=(X^3*Y^3*Z^3)^(1/3)=XYZ
\n" ); document.write( "HENCE
\n" ); document.write( " x^2y+y^2z+z^2x>=3xyz \r
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