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.

Hint: let $x=a+b-c$, $y=a+c-b$, $z=b+c-a$. Clearly
$$a=\frac{x+y}2,\ b=\frac{x+z}{2},\ c=\frac{y+z}{2},\ a+b=\frac{2x+y+z}{2},\cdots$$
hence the original inequality reads
$$\frac{x+y}{x+y+2z}+ \frac{x+z}{x+2y+z}+\frac{y+z}{2x+y+z}<2.$$
Note that
$$\frac{x+y}{x+y+2z}+ \frac{x+z}{x+2y+z}+\frac{y+z}{2x+y+z}<\frac{x+y}{x+y+z}+
\frac{x+z}{x+y+z}+\frac{y+z}{x+y+z}\cdots$$
