Question

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

• A. $1$
• B. $2$
• C. $3$
• D. $6$

Answer 1

Answer: (D)

$6$

let $\frac{b+c}{a},\frac{c+a}{b},\frac{a+b}{c}$ are three positive number, ($a,b,c$ are positive real numbers)

applying property, $A.M.\ge G.M.$

$\frac{\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}}{3}\ge (\frac{1}{abc}(a+b)(b+c)(c+a){)}^{\frac{1}{3}}$ .......(1)

Applying the same for two numbers $a,b$ and $b,c$ and $c,a$

$(a+b)\ge 2\sqrt{ab}$

$(b+c)\ge 2\sqrt{bc}$
$(c+a)\ge 2\sqrt{ac}$

from these three we can write,

$(a+b)(b+c)(c+a)\ge 8abc$

putting this value in RHS of eq (1)

minimumum value of $G.M.=2$

hence from eq (1)

$\frac{\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}}{3}\ge 2$

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