Question

If $a,b,c$ are the sides of a triangle, then
$\frac{a}{b+c-a}+\frac{b}{c+a-b}+\frac{c}{a+b-c}$

• A. $\le 3$
• B. $\ge 3$
• C. $\le 2$
• D. $\ge 2$

Answer 1

The correct option is B $\ge 3$

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

let $b+c-a=x$

$c+a-b=y$ and

$a+b-c=z$

Therefore, $x+y=b+c-a+c+a-b⟹x+y=2c⟹c=\frac{x+y}{2}$

$y+z=c+a-b+a+b-c⟹y+z=2a⟹a=\frac{y+z}{2}$ and

$x+z=b+c-a+a+b-c⟹x+z=2b⟹b=\frac{x+z}{2}$

substituting the above expressions in the given equation we get

$\frac{a}{b+c-a}+\frac{b}{c+a-b}+\frac{c}{a+b-c}=\frac{y+z}{2x}+\frac{x+z}{2y}+\frac{x+y}{2z}$

multiplying 2 on both sides

$2(\frac{a}{b+c-a}+\frac{b}{c+a-b}+\frac{c}{a+b-c})=\frac{y+z}{x}+\frac{x+z}{y}+\frac{x+y}{z}$

$2(\frac{a}{b+c-a}+\frac{b}{c+a-b}+\frac{c}{a+b-c})\ge 6$

since, $\frac{y}{x}+\frac{z}{x}+\frac{x}{y}+\frac{z}{y}+\frac{x}{z}+\frac{y}{z}\ge 6$

Therefore,

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