Question

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

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

Answer 1

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

Since, $a,b$ and $c$ are in AP.
Let $d$ be the common difference.
$\therefore a=b-d,b=d,c=b+d$
Also, $abc=4$
$\Rightarrow (b-d)d(b+d)=4$
$\Rightarrow (b^2-d^2)b=4$
$\Rightarrow b^3=4+d^{2}b$
$\Rightarrow b^3\ge 4$
$\Rightarrow b\ge {\left(2\right)}^{2/3}$