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: We suppose $a \neq 0$, otherwise it is not a quadratic equation.
$$ax^2+bx+c>0,\forall x \in \mathbb{R} \Rightarrow ax^2+bx+c\neq0,\forall x \in \mathbb{R}\Rightarrow D<0$$
Also $\lim_{x\to \infty}ax^2+bx+c=\frac{a}{|a|}\infty$. So :
$$ax^2+bx+c>0,\forall x \in \mathbb{R} \Rightarrow \lim_{x\to \infty}ax^2+bx+c =+\infty \Rightarrow \frac{a}{|a|}=1 \Rightarrow a>0$$
Finally : $$ax^2+bx+c>0,\forall x \in \mathbb{R} \Rightarrow D<0 \space \text{and} \space a>0$$
You can do the same kind of reasonning for the second inequality.
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.
