Question

If three positive real numbers a, b, c are in A.P. and abc = 4, then the minimum possible value of b is_________.

Answer 1

Let us suppose three positive real numbers a, b and c are in A.P and abc = 4
Since arithmetic mean ≥ geometric mean
$$\begin{gathered} \mathrm{i}.\mathrm{e}\frac{a+b+c}{3}\ge \sqrt[3]{abc} \\ \mathrm{i}.\mathrm{e}\frac{a+b+c}{3}\ge \sqrt[3]{4}\left(\mathrm{given}\right) \\ \mathrm{Since}b=\frac{a+c}{2} \\ 2b=a+c \\ \mathrm{i}.\mathrm{e}\frac{2b+b}{3}\ge {\left(4\right)}^{\frac{1}{3}} \\ \mathrm{i}.\mathrm{e}\frac{3b}{3}\ge {\left(4\right)}^{\frac{1}{3}} \\ \mathrm{i}.\mathrm{e}b\ge {\left(2\right)}^{\frac{2}{3}} \end{gathered}$$

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