Question

If $a,b,c$ are positive such that $a{b}^2{c}^3=64$ then least value of $(\frac{1}{a}+\frac{2}{b}+\frac{3}{c})$ is

• A. 6
• B. 2
• C. 3
• D. 32

Answer 1

Answer: (C)

3

Given numbers $a,b,c$ such that $a{b}^2{c}^3=64$
Since,$G.M.\ge H.M.$
$(a.b.b.c.c.c{)}^{\frac{1}{6}}\ge \frac{6}{\frac{1}{a}+\frac{1}{b}+\frac{1}{b}+\frac{1}{c}+\frac{1}{c}+\frac{1}{c}}$
$\Rightarrow (64{)}^{1/6}\ge \frac{6}{\frac{1}{a}+\frac{2}{b}+\frac{3}{c}}$
$\Rightarrow \frac{1}{a}+\frac{2}{b}+\frac{3}{c}\ge 3$