Question

If a, b, c $>0$ and abc $=8$, then the minimum value of $\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}$.

Answer 1

Given $abc=8$

We know that A.M $\ge$G.M for finite number of real terms

Consider teh terms to be $\frac{ab}{c},\frac{bc}{a},\frac{ca}{b}$

A.M $=\frac{\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}}{3}$

G.M $=\sqrt[3]{3\frac{ab}{c}.\frac{bc}{a}.\frac{ca}{b}}=\sqrt[3]{3abc}=2$

A.M $\ge$G.M

$\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}\ge 6$

the minimum value is $6$