Inequality of sides of triangle If a,b,c are the sides of a triangle ABC then prove $$ a/(b+c) + b/(a+c) + c/(a+b) < 2 $$I tried to solve this by triangle inequality but i was not able to get to the solution.

As $a,b,c>0$ $$ a/(b+c) + b/(a+c) + c/(a+b) - 2 =\frac{\sum a^2(b+c-a)}{(b+c)(c+a)(a+b)}$$ which is $>0$ as $b+c>a$ etc.
