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Question

Suppose z>y>x>0 and S=x2+y2+z2x+y+z, then


  1. S>x2z
     

  2. S<z2x
     

  3. S,y2x
     

  4. none of these


Solution

The correct options are
A

S>x2z
 


B

S<z2x
 


Since z>y>x>0, we have z2>y2>x2.
Therefore,
3x2<x2+y2+z2<3z2             (1)and           3x<x+y+z<3z            (2)From(2), we get13z<1x+y+z<13x3x23z<x2+y2+z2x+y+z<3z23xx2z<x2+y2+z2x+y+z<z2x

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