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 !

A standard trick to solve such problems is to use the substitution $a= y + z, b =x+z, c = x+y$, where $x, y, z>0$ (to see why this is so, draw the inscribed circle and pairs of equal tangents: $x, y, z$ are the lengths of the tangents)

So, we have $$x+y+z = \frac{1}{2}$$
and the inequality is $$ (x+y)^2 + (x+z)^2 + (y+z)^2  + 4 (x+y)(x+z)(y+z) < x+y+z$$
$$2x^2 + 2y^2 + 2z^2 + 2(xy+yz+xz) + 4 (x+y)(x+z)(y+z) < x+y+z $$
$$x^2 + y^2 +z^2 + (xy+yz+xz) + 2(x+y)(x+z)(y+z) < (x+y+z)^2 $$
$$ 2(x+y)(x+z)(y+z) < xy + xz +yz$$
$$(x+y)(x+z)(y+z) < (x+y+z) (xy+xz +yz)$$
$$ x^2y + xzy + x^2z + xz^2  + y^2x + y^2z + yxz + yz^2  < x^2y + x^2z + xyz + xy^2 +xyz +y^2z + xyz +xz^2 + yz^2 $$
$$2 xyz < 3xyz $$
