Question

For positive real numbers a, b, and c, the minimum value of $\frac{b+c}{a},\frac{c+a}{b},\frac{a+b}{c}$ is

• A. $5$
• B. $6$
• C. $9$
• D. None of these

Answer 1

Answer: (B)

$6$

For $3$ positive real number a,b,c

$AM=\frac{a+b+c}{3},GM=\sqrt[3]{3abc}$

$HM=\frac{3abc}{ab+bc+ca}=\frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}$

We know $HM\le GM\le AM$

So $\frac{a+b+c}{3}\ge \frac{3abc}{ab+bc+ca}$

$\frac{(ab+bc+ca)(a+b+c)}{abc}\ge 9$

$(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\ge 9$

For minimum $(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})=9$

Now $\frac{b+c}{a}+\frac{c+a}{b}+\frac{a+b}{c}$

$\frac{b+c+a-a}{a}+\frac{c+a+b-b}{b}+\frac{a+b+c-c}{c}$

$=(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})-3$
$=9-3=6$

Option B