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 !

Homogenizing the equation, we WTS
$$\frac{(a+b+c)^3}{2} -(a+b+c)(a^2+b^2+c^2)-4abc > 0$$
Expanding and factoring, we obtain
$$(a+b-c)(a-b+c)(-a+b+c) > 0$$
This is true via the triangle inequality.
