solve this ineqalite<excuse my english>?

Question

when a,b,c are the lenghts of an triangle then:
a/(b+c-a)+ b/(c+a-b)+(a+b-c) > or = 3

Answer 1

I'm taking the liberty of assuming you meant:

a/(b+c-a)+ b/(c+a-b)+c/(a+b-c) >= 3

The three denominators are never zero so we can multiply through by their product to get the equivalent inequality:

a(c+a-b)(a+b-c)+ b(b+c-a)(a+b-c) + c(b+c-a)(c+a-b)
- 3(b+c-a)(c+a-b)(a+b-c) >= 0

The original inequality will be true if and only if this one is. Expanding the left side and dividing by 2 gives:

6cba + 2c^3 + 2b^3 + 2a^3 - 2bc^2 - 2ac^2 - 2cb^2 - 2ab^2 - 2ca^2 - 2ba^2

We only need to prove that this is nonnegative. This can be done by inspecting the following expression and expanding it to show that it gives the same expression:

(a+b-c)(a-b)^2 + (a+c-b)(a-c)^2 + (b+c-a)(b-c)^2

Since each of the factors in the above expression is positive (a+b-c > 0 etc., by the triangle inequality) it must be nonnegative.
QED

Moreover, it's an equation if and only if all three terms are zero (i.e. a=b=c). In other words the inequality is strict unless the triangle is equilateral.