Question

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

Given:
$a,b,c$ are in A.P.
$abc=4$
As $a,b,c>0$,
Using $A.M.\ge G.M.$
$\Rightarrow \frac{a+b+c}{3}\ge (abc{)}^{1/3}\Rightarrow \frac{3b}{3}\ge 4^{1/3}(\because a+c=2b)\Rightarrow b\ge 2^{2/3}$

Hence, the minimum value of $b$ is $2^{2/3}$