show that a/(c+a-b)+b/(a+b-c)+c/(c+b-a)>=3.if a,b,c are the sides of triangle

As we know, in any triangle, the sum of any two sides is greater than the third side.

So, the denominator of all the fractions in the above expression would be positive. Also, if the triangle is an equilateral one, then the value of the given expression would be 3 (and that's the minimum value).

In any other case, the value would exceed 3. You can check this by taking random values for three sides of a triangle-  for instance (2,4,6) or (3,4,5).