If three positive real numbers a, b, c are in A.P. and abc = 4,

Let us suppose three positive real numbers a, b and c are in A.P and abc = 4

Since arithmetic mean ≥ geometric mean

i. e $\frac{a+b+c}{3} \geq \sqrt[3]{a b c}$

i. e $\frac{a+b+c}{3} \geq \sqrt[3]{4}$   (given)

Since $b=\frac{a+c}{2}$

$2 b=a+c$

i. e $\frac{2 b+b}{3} \geq(4)^{\frac{1}{3}}$

i.e $\frac{3 b}{3} \geq(4)^{\frac{1}{3}}$

i. e $b \geq(2)^{\frac{2}{3}}$

i.e minimum possible value of $b$ is $(2)^{\frac{2}{3}}$.