inequality $(a+c)(a+b+c)<0$, prove $(b-c)^2>4a(a+b+c)$ 
If $(a+c)(a+b+c)<0,$
  prove $$(b-c)^2>4a(a+b+c)$$

I will use the constructor method that want to know can not directly prove it?

Because
$$(b-c)^2-4a(a+b+c)=(b-c)^2+(4a+8c)(a+b+c)-8(a+c)(a+b+c)=$$
$$=(2a+b+3c)^2-8(a+c)(a+b+c)>0$$
