Question

If $a,b$ and $c$ are $3$ positive real numbers, then minimum value of $\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+b}{a}$ is

Answer 1

$\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}=\frac{a}{c}+\frac{b}{c}+\frac{b}{a}+\frac{c}{a}+\frac{c}{b}+\frac{c}{a}=(\frac{a}{c}+\frac{c}{a})+(\frac{a}{b}+\frac{b}{a})+(\frac{b}{c}+\frac{c}{b})A.M.\ge G.M.\frac{\frac{a}{b}+\frac{b}{a}}{2}\ge \sqrt{\frac{a}{b}.\frac{b}{a}}$(a,b,c are positive)

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

Similarly, $\frac{b}{c}+\frac{c}{b}\ge 2,\frac{a}{c}+\frac{c}{a}\ge 2\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}=(\frac{a}{c}+\frac{c}{a})+(\frac{a}{b}+\frac{b}{a})+(\frac{b}{c}+\frac{c}{b})\ge 6$

Minimum value of $\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}=6$