If three positive real numbers x,y,z satisfy y-x=z-y and xyz=4, then what is the minimum possible value of y?
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Note that x , y , z forms an arithmetic progression. Without the loss of generality, assume x ≤ y ≤ z . Let x , y , z be x , x + a , x + 2 a for some nonnegative real a, respectively.
By AM-GM inequality, ( 3 x + ( x + a ) + ( x + 2 a ) ) 3 ≥ x ( x + a ) ( x + 2 a ) = 4 .
Therefore, y = x + a ≥ 2 3 2 . Equality occurs when a = 0 ( x = y = z )