let a,b,c be the sides of a triangle and $a+b+c=2$ then prove that $ a^2 + b^2 + c^2 +2abc < 2$ let a,b,c be the sides of a triangle and $a+b+c=2$ then prove that
 $ a^2 + b^2 + c^2 +2abc < 2$
I used the fact that $   a+b-c>0  $ and multiplied the 3 cyclic equations but coudn't reach the final equation.

Just continue your approach, focusing on reaching the goal: 
$$(a+b-c)(b+c-a)(c+a-b) > 0 \iff (1-c)(1-a)(1-b) > 0 $$
$$\iff 1-(a+b+c)+(ab+bc+ca)-abc > 0$$
$$\iff 1-(2)+\tfrac12 \left((a+b+c)^2-a^2-b^2-c^2 \right)-abc > 0$$
$$\iff -2+ \left(4-a^2-b^2-c^2 \right)-2abc > 0$$
$$\iff a^2+b^2+c^2 + 2abc < 2$$
