Question

If three positive numbers $a,b$ and $c$ are in $A.P.$ such that $abc=8$, then the minimum possible value of $b$ is

• A. $2$
• B. $4^{\frac{1}{3}}$
• C. $4^{\frac{2}{3}}$
• D. $4$

Answer 1

Answer: (A)

$2$

As $a,b,c$ are in A.P.....given

$\therefore$ $b=\frac{a+c}{2}$

For three numbers, Arithmetic mean $\ge$ Geometric mean

$\Rightarrow \frac{a+b+c}{3}\ge (abc{)}^{\frac{1}{3}}$

$\Rightarrow$Since $a,b,c$ are in A.P., we have $\frac{a+b+c}{3}=b$

$\Rightarrow \frac{2b+b}{3}\ge (abc{)}^{\frac{1}{3}}$

$\Rightarrow b\ge (abc{)}^{\frac{1}{3}}$

$\therefore b\ge 2$ since $abc=8$