prove that : $a^{2}+b^{2}+c^{2}+4abc<\frac12$ Let $a,b,c$ be a sides of triangle 
Such that : $a+b+c=1$ Then 

prove that : $a^{2}+b^{2}+c^{2}+4abc<\frac{1}{2}$

My effort :
Since $a+b+c=1$ $\implies$ $2S=sr=bc\sin A=\frac{abc}{2R}$
Also : $S=\sqrt{s(s-a)(s-b)(s-c)}$ 
Also : 
$a^{2}+b^{2}+c^{2}+2(ab+ac+bc)=1$
But I don't know how to complete it,any help is appreciated !

Using homogeneization and Ravi's substitution, the problem boils down to showing that
$$2(a+b+c)(a^2+b^2+c^2)+8abc \leq (a+b+c)^3 $$
holds for any triple $(a,b,c)$ of side lengths of a triangle, i.e. that
$$4(A+B+C)((B+C)^2+(A+C)^2+(A+B)^2)+8(A+B)(A+C)(B+C) \leq 8(A+B+C)^3 $$
holds for any triple $(A,B,C)$ of positive numbers. The previous inequality simplifies into
$$ (A+B+C)(A^2+B^2+C^2+AB+AC+BC)+(A^2 B+B^2 A+A^2 C+C^2 A+B^2 C+C^2 B+2ABC) \leq (A+B+C)^3$$
which is equivalent to the trivial $-ABC\leq 0$.
