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?
 A: Consider the quadratic
$$ f(x) = ax^2 - (b-c)x + (a+b+c) $$
$$f(1)f(0) = 2(a+c)(a+b+c) \lt 0$$
Thus if $a \neq 0$, then this has a real root in $(0,1)$ and so
$$(b-c)^2 \ge 4a(a+b+c)$$
If $(b-c)^2 = 4a(a+b+c)$, then we have a double root in $(0,1)$ in which case, $f(0)$ and $f(1)$ will have the same sign.
Thus $$(b-c)^2 \gt 4a(a+b+c)$$
If $a = 0$, then $c(b+c) \lt 0$, and so we cannot have $b=c$ and thus $(b-c)^2 \gt 0 = 4a(a+b+c)$
And if you want a more "direct" approach, we show that $(p+q+r)r \lt 0 \implies q^2 \gt  4pr$ using the following identity:

$$(p+q+r)r = \left(p\left(1 + \frac{q}{2p}\right)\frac{q}{2p}\right)^2 + \left(r - \frac{q^2}{4p}\right)^2 + p\left(r - \frac{q^2}{4p}\right)\left(\left(1 + \frac{q}{2p}\right)^2 + \left(\frac{q}{2p}\right)^2\right)$$

If $(p+q+r)r \lt 0$, then we must have have $p\left(r - \frac{q^2}{4p}\right) \lt 0$, as all the other terms on the right side are non-negative.
Of course, this was gotten by completing the square in $px^2 + qx + r$ and setting $x=0$ and $x=1$ and multiplying.
A: Because
$$(b-c)^2-4a(a+b+c)=(b-c)^2+(4a+8c)(a+b+c)-8(a+c)(a+b+c)=$$
$$=(2a+b+3c)^2-8(a+c)(a+b+c)>0$$
