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 $2$.

• A. True
• B. False

Answer 1

The correct option is A True

We know for any triangle,
$a+b>c=>\frac{c}{a+b}<1$
similarly $\frac{a}{b+c}<1$ and $\frac{b}{c+a}<1$
Adding all we get, $\frac{c}{a+b}+\frac{a}{b+c}+\frac{b}{c+a}<3$
Hence given statement is true.