Question

If $a,b,c$ are positive real numbers such that $a{b}^2{c}^3=64$ then minimum value of $(\frac{1}{a}+\frac{2}{b}+\frac{3}{c})$ is equal to.

• A. $6$
• B. $3$
• C. $2$
• D. $Noneofthese$

Answer 1

The correct option is B $3$

Using, $A.M\ge G.M$
$\frac{(\frac{1}{a}+\frac{2}{b}+\frac{3}{c})}{6}\ge {\left(\frac{1}{a}\frac{1}{b^2}\frac{1}{c^3}\right)}^6=\frac{1}{2}$
$\Rightarrow (\frac{1}{a}+\frac{2}{b}+\frac{3}{c})\ge 3$
Hence, option 'B' is correct.