If three positive real numbers a,b,c  are AP such that abc=4,  then the minimum possible value of b  is

Let d be the common difference of the AP, then         4=abc=(b−d)b(b+d)=b(b2−d2)⇒     b3-bd2=4 ⇒ b3 =4+bd2 ⇒ b3 ≥4                                       [∵  b>0,d2≥0]⇒     b≥22/3Thus, minimum possible value of b is 22/3, that is the case when  d=0 .