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

Sequences and Series

Solution:

G.M. $\geq$ H.M.
$(a.b.b.c.c.c) ^{\frac{1}{6}} \geq \frac{6}{\frac{1}{ a }+\frac{1}{ b }+\frac{1}{ b }+\frac{1}{ c }+\frac{1}{ c }+\frac{1}{ c }}$
$\Rightarrow(64)^{1 / 6} \geq \frac{6}{\frac{1}{ a }+\frac{2}{ b }+\frac{3}{ c }} $
$\Rightarrow \frac{1}{ a }+\frac{2}{ b }+\frac{3}{ c } \geq 3$