Question

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

Answer 1

it will be greater than or equal to 3/2

proof To prove this we need first to show that the least value of the lefthand side occurs when a/(b+c) = b/(c+a) = c/(a+b). If a+b+c = constant and if we assume that c is fixed, then a+b is fixed but we vary a and b to minimize the lefthand side, so c/(a+b) = constant and we consider the sum of the other two fractions. a/(b+c) + b/(c+a) a(c+a) + b(b+c) a^2 + b^2 + c(a+b) = -------------- = ------------------- (b+c)(c+a) (b+c)(c+a) (a+b)^2 - 2ab + c(a+b) = ----------------------- remember (b+c)(c+a) a+b is fixed, the top line is a minimum if 2ab is a maximum, and the bottom line is a maximum if (b+c)(c+a) is a maximum, i.e. if the two factors are equal. Both these conditions require a = b. With a = b, if we now allow c to vary, a similar argument leads to c = b, and so for the least value on the lefthand side we require a = b = c. Then with a = b = c we obtain the value of the lefthand side: 1/2 + 1/2 + 1/2 = 3/2 and this is the minimum value. It follows that a/(b+c) + b/(c+a) + c/(a+b) >= 3/2