Question

If $a,b,c$ are the sides of a triangle, then the minimum value of $\frac{a}{b+c-a}+\frac{b}{c+a-b}+\frac{c}{a+b-c}$ is equal to

• A. 3
• B. 6
• C. 9
• D. 12

Answer 1

The correct option is A 3

Given expression

$\frac{1}{2}\sum \frac{2a}{b+c-a}=\frac{1}{2}\sum (\frac{2a}{b+c-a}+1)-\frac{3}{2}$

$=\frac{1}{2}(a+b+c)\sum \left(\frac{1}{b+c-a}\right)-\frac{3}{2}$

Now, as $(a+b+c)=\sum (b+c-a)$

Applying $A.M\ge H.M.$

Minimum value of the expression $=\frac{1}{2}\times 9-\frac{3}{2}=3$