Let x , y , z be the positive real numbers such that x y z = 1 . Find the maximum value of A = ( x + 1 ) 2 + y 2 + 1 1 + ( y + 1 ) 2 + z 2 + 1 1 + ( z + 1 ) 2 + x 2 + 1 1
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We notice that:
( x + 1 ) 2 + y 2 + 1 = x 2 + y 2 + 2 x + 2 = ( x − y ) 2 + 2 ( x y + x + 1 ) ≥ 2 ( x y + x + 1 )
and x , y > 0
⟹ ( x + 1 ) 2 + y 2 + 1 1 ≤ 2 ( x y + x + 1 ) 1
Similarly, ( y + 1 ) 2 + z 2 + 1 1 ≤ 2 ( y z + y + 1 ) 1 , ( z + 1 ) 2 + x 2 + 1 1 ≤ 2 ( z x + z + 1 ) 1 .
So, A ≤ 2 1 ( x y + x + 1 1 + y z + y + 1 1 + z x + z + 1 1 ) = 2 1
(by applying the result of this problem )
When x = y = z = 1 we have A = 1 / 2 , so A m i n = 2 1