Question

If three positive real numbers $a,b,c$ are in $A.P.$ such that $abc=4$, 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 $2^{2/3}$

Let the number be $p-d,p,p+d$ which are in A.P. with the first term as $p-d$ and common difference as $d$.
Given, $abc=4\Rightarrow (p-d)\left(p\right)(p+d)=4$

$\Rightarrow p(p^2-d^2)=4$
As all $3$ numbers are positive, we can use AM$\ge$GM inequality.
$\frac{p-d+p+p+d}{3}\ge {\left(\right(p-d\left)\right(p\left)\right(p+d\left)\right)}^{\frac{1}{3}}$
$\Rightarrow p^3\ge p(p^2-d^2)$

$\Rightarrow p\ge 4^{\frac{1}{3}}$
Hence, the minimum possible value of $p$ or $b$ is $4^{\frac{1}{3}}$ or $2^{\frac{2}{3}}$

Hence, option B is the correct answer.