Question

If $a,b$ and $c$ represent the lengths of sides of a triangle, then the possible integral value of $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}$ is

Answer 1

We know, $0<a<b+c\Rightarrow \frac{a}{b+c}<1$(Triangle Inequality)
$\therefore \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}<3$
In the case of $a=b=c$, we have $\frac{a}{b+c}=\frac{1}{2}$
$\Rightarrow \frac{3}{2}\le \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}<3$
$\therefore$ Integral values of $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=2$