Question

If a,b,c be positive, then minimum value of $\frac{b+c}{a}$ + $\frac{c+a}{b}$ + $\frac{a+b}{c}$ is __

Answer 1

Since Arithmetic mean $\ge$ Geometric mean

$\frac{\frac{a}{b}+\frac{b}{a}}{2}$ $\ge$ $\sqrt{\frac{a}{b}.\frac{b}{a}}$

$\Rightarrow$ $\frac{a}{b}+\frac{b}{a}$ $\ge$ 2 -----------------(1)

Similarly, $\frac{b}{c}+\frac{c}{b}$ $\ge$ 2 ------------------(2)

and $\frac{c}{a}+\frac{a}{c}$ $\ge$ 2 -------------------(3)

Adding the corresponding sides of (1),(2) and (3),

$(\frac{a}{b}+\frac{b}{a})$ + $(\frac{b}{c}+\frac{c}{b})$ + $(\frac{c}{a}+\frac{a}{c})$ $\ge$ 6

(or) $(\frac{b}{a}+\frac{c}{a})$ + $(\frac{c}{b}+\frac{a}{b})$ + $(\frac{a}{c}+\frac{b}{c})$ $\ge$ 6

$\Rightarrow$ $\frac{b+c}{a}$ + $\frac{c+a}{b}$ + $\frac{a+b}{c}$ $\ge$ 6

So, the minimum value is 6.