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Problem 2.
Prove that for all nonnegative real numbers x,y and z, which are not all equal to 0, the following inequality holds:  Determine all triples (x,y,z) for which the equality holds.
Solution
It is written:   , (*)Let x+y2+z2=a,
y+z2+x2=b,
z+x2+y2=c
inequality (*) reduces to AM-HM inequality
 Equality holds for a=b=c
 The above is equivalent to  , from which we deduce all cases of equality, which are:1)  ,2)  , 3)  , 0 ≤ y ≤ 14)  . 0 ≤ x ≤ 1
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