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

         $4=abc=\left(b-d\right)b\left(b+d\right)=b\left(b^2-d^2\right)$

$\Rightarrow b^3-b{d}^2=4$ $\Rightarrow b^{3 }=4+b{d}^2$ $\Rightarrow b^3 \ge 4$                                       $\left[\because b>0,d^2\ge 0\right]$

$\Rightarrow b\ge 2^{2/3}$

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