Question

If three positive real numbers $\mathrm{a},\mathrm{b}$ and $\mathrm{c}$ are in AP and $\mathrm{abc}=4$, then the minimum possible value of $\mathrm{b}$ is

• A. $2^{\frac{3}{2}}$
• B. $2^{\frac{2}{3}}$
• C. $2^{\frac{1}{3}}$
• D. $2^{\frac{5}{2}}$

Answer 1

The correct option is B

$2^{\frac{2}{3}}$

Explanation for the correct option:

Finding the minimum value of $\mathbf{b}$:

Since, $\mathrm{a},$$\mathrm{b}$ and $\mathrm{c}$ are in A.P then,

$2\mathrm{b}=\mathrm{a}+\mathrm{c}...\left(1\right)$

We know, $\mathrm{AM}\ge \mathrm{GM}$, for positive quantities, then

$$\begin{gathered} \left(\frac{\mathrm{a}+\mathrm{b}+\mathrm{c}}{3}\right)\ge \sqrt[3]{\mathrm{abc}} \\ \Rightarrow \frac{2b+\mathrm{b}}{3}\ge \sqrt[3]{\mathrm{abc}}\mathbf{[}\mathbf{\because }\mathbf{}\mathbf{from}\mathbf{}\mathbf{1}\mathbf{]} \\ \Rightarrow \mathrm{b}\ge \sqrt[3]{4} \\ \Rightarrow \mathrm{b}\ge {\left(2^2\right)}^{\frac{1}{3}} \\ \Rightarrow \mathrm{b}\ge {\left(2\right)}^{\frac{2}{3}} \end{gathered}$$

Hence, the correct option is B.