If a, b, c are the sides of a triangle, then a/(b+c-a) + b/(c+a-b)+ c/(a+b-c) is

Question

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

(a) < 3 

(b) ≥ 2 

(c) ≥ 3 

(d) < 2

Answer 1

(c) > 3

Since a, b, c are the sides of a triangle, 

a + b – c > 0, b + c – a > 0, c + a – b > 0 

Let x = b + c – a, y = c + a – b, z = a + b – c 

Then x + y = 2c, y + z = 2a, z + x = 2b.