Difficulty in understanding the CBS-inequality proof. 

I can't understand why the discriminant has to be negative here. Could someone please elaborate?
 A: Consider the polynomial $f(x) = Ax^2 +B x+ C$. There are $3$ cases for the discriminant.


*

*$D>0 \iff 2 \text { distinct real roots } x_1, x_2, \text { with } x_1 < x_2$. The sign of $f$ is the same as the sign of $A$ on the interval $(-\infty, x_1)\cup (x_2,\infty)$, while the sign of $f$ is the opposite of the sign of $A$ on the interval $(x_1,x_2).$

*$D = 0 \iff 1 \text{ real root } x_0 .$ The sign of $f$ is the same as the sign of $A$ on the interval $(-\infty, x_0) \cup (x_0, \infty)$.

*$D< 0 \iff \text { no real roots }$. The sign of $f$ is the same as the sign of $A$ on $\mathbb R$.

$f(x) \begin{array}[t]{l}= Ax^2 + Bx + C = A \left(x^2+\frac BA x +\frac CA\right)=A\left(x^2  +2\cdot \frac BA x +\frac{B^2}{4A}-\frac{B^2}{4A^2}+\frac{4AC}{4A^2} \right)\\
=A\big[\left(x+\frac{B}{A}\right)^2-\frac{B^2-4AC}{4A^2}\big]=A\big[\left(x+\frac BA\right)^2-\frac{D}{4A^2}\big].
\end{array}$
Thus, the sign of $f$ basically depends on $D$, since $\left (x+\frac BA\right)^2\ge 0, \forall x\in\mathbb R.$


*

*$D = 0 \iff f(x) = A\cdot \underbrace{\left(x+\frac BA\right)^2}_{ \ge 0}$ 

*$D < 0 \iff f(x) = A\cdot \underbrace{\big[\left(x+\frac BA\right)^2-\frac{D}{4A^2}\big]}_{>0}$
