if 3 positive nos, x,y,z are in AP and xyz=4.......then find the minimum possible value of y... PLS EXPLAIN...!!>>WHATS THE LOGIC BEHIND THIS...???? PLS ELABORATE...answer will surely be approved..

2^2/3 is the answer

firstly differentiate and equate the equation to zero

x,y and z are in AP
so, 2y = x+z

Now by

AM>= GM
so,(x+y+z)/3 >= cube root (x*y*z)

So,3y/3>=cube root(x*y*z)
=>y>=4^1/3
=>y>= 2^2/3
So,minimum value of y = 2^2/3

approve it plz