Question

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

• A. $2^{1/3}$
• B. $2^{2/3}$
• C. $2^{1/2}$
• D. $2^{3/2}$

Answer 1

The correct option is B $2^{2/3}$

We know, from $AM\ge GM,$ for positive quantities, that

$\frac{(a+b+c)}{3}\ge \sqrt[3]{3abc}\frac{(a+c)+b}{3}\ge \sqrt[3]{3abc}$

But $2b=(a+c).$

This implies $\frac{3b}{3}\ge {\left(4\right)}^{1/3}b\ge {\left(2\right)}^{2/3}$