<s>[INST] 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.
 [/INST] Let's homogenize the equation by multiplying with $ a+ b + c = 2$.
We need to show that
$$ 2 ( a + b + c) (a^2 + b^2 + c^2) + 8abc < (a+b+c)^3.$$
Expanding this, it is equivalent to showing that 
$$ (a+b-c) ( b+c-a) ( c+a-b) > 0. $$
Hence we are done.
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