If three positive real numbers a,b,c are in A.P. such that $abc=4$ then the minimum possible value b is

Let d be the common difference of the AP. then $4=abc=(b-d)b(b+d)=b(b^2-d^2)\Rightarrow b^3=4+b{d}^2\ge 4[\because b>0,d^2\ge 0]\Rightarrow b\ge 2^{2/3}$ 

Thus, minimum possible value of b is$2^{2/3}$  , that is the case when d = 0.