Q:

Quadratic inequality proof.

I recently encountered a quadratic equations property that 
$ax^2+bx+c>0$  $  \forall$   $x\in \Re \Rightarrow D<0$ $and$ $a>0$ 
and $ax^2+bx+c<0$  $  \forall$   $x\in \Re \Rightarrow D<0$ $and$ $a<0$.
Now, i tried to prove it algebraically and through graphs. Its clear that the equation seems much simple by just making a parabola and as $D<0$ the parabola never touches the $x$ axis and hence the two equations follow.
Can there be any algebraic proof the relation mentioned above. 

A:

For $a>0$ we have:
$$
ax^2+bx+c>0 \quad \iff \quad x^2+\frac{b}{a}+\frac{c}{a}>0
$$
so:
$$
x^2+\frac{b}{a}>-\frac{c}{a}
$$
$$
x^2+\frac{b}{a} +\left(\frac{b}{2a} \right)^2>-\frac{c}{a}+\left(\frac{b}{2a} \right)^2
$$
$$
\left(x+\frac{b}{2a} \right)^2>-\frac{c}{a}+\left(\frac{b}{2a} \right)^2
$$
that is always verified ( since The LHS is a square) if
$$
-\frac{c}{a}+\left(\frac{b}{2a} \right)^2<0
$$
i.e.
$$
\frac{b^2-4ac}{4a^2}=\frac{\Delta}{4a^2} <0
$$
and since $4a^2>0$ we have $\Delta<0$.
You can do the same for the other case.

