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

Since a,b, c are in A.P., therefore, b -a = d and c -b = d,
where dis the common difference of the A.P.
$\therefore \mathrm{a}=\mathrm{b}-\mathrm{d}\text{ and }\mathrm{c}=\mathrm{b}+\mathrm{d}$
Now, abc=4
$\begin{array}{l}\Rightarrow (\mathrm{b}-\mathrm{d})\mathrm{b}(\mathrm{b}+\mathrm{d})=4\\ \Rightarrow \mathrm{b}\left({\mathrm{b}}^2-{\mathrm{d}}^2\right)=4\end{array}$
But, $\mathrm{b}\left({\mathrm{b}}^2-{\mathrm{d}}^2\right)<\mathrm{b}\times {\mathrm{b}}^2$
$\begin{array}{l}\Rightarrow \mathrm{b}\left({\mathrm{b}}^2-{\mathrm{d}}^2\right)<{\mathrm{b}}^3\\ \Rightarrow 4<{\mathrm{b}}^3\\ \Rightarrow {\mathrm{b}}^3>4\\ \Rightarrow \mathrm{b}>2^{2/3}\end{array}$
Hence, the minimum value of b is  2 ^2/3