On Cauchy - Schwarz Inequality: quadratic polynomial. Denote $\mathbb{F}=\mathbb{C}$ or $\mathbb{R}$.
Theorem (Cauchy - Schwarz Inequality). If $\langle\cdot,\cdot\rangle$ is a semi-inner product on a vector space $H$, then $$\lvert\langle x,y\rangle\rvert\le\lVert x\rVert\lVert y\rVert,\quad\textit{for all}\;x,y,\in H.$$
Proof. If $x=0$ or $y=0$, then there is nothing to prove, so suppose that $x$ and $y$ are both nonzero.
Given any scalar $z\in\mathbb{F}$, there is a scalar $\alpha$ with modulus $\lvert\alpha\rvert=1$ such that $\alpha z=\lvert z \rvert$. In particular, if we set $z=\langle x, y\rangle$, then there is a scalar with $\lvert \alpha \rvert=1$ such that $$\langle x,y\rangle=\alpha\lvert\langle x,y\rangle\rvert.$$ Multiplying both sides by $\overline{\alpha}$, we see that we also have $\overline{\alpha}\langle x, y\rangle=\lvert\langle x, y \rangle \rvert$.
For each $t\in\mathbb{R}$, using the Polar Identity and antilinearity in the second variable, we compute that
\begin{equation}
\begin{split}
0\le\lVert x-\alpha ty\rVert= &\rVert x\lVert^2-2\Re\big(\langle x,\alpha ty\rangle\big)+t^2\rVert y \lVert^2 \\
= &\rVert x\lVert^2-2t\Re\big(\overline{\alpha}\langle x, y\rangle\big)+t^2\lVert y \lVert^2\\ 
= & \lVert x \rVert ^2-2t\lvert\langle x, y\rangle\rvert+t^2\lVert y \rVert^2\\
 = & at^2+bt+c, 
\end{split}
\end{equation}
where $a=\lVert y \rVert ^2$, $b=-2\lvert \langle x, y\rangle \rvert$, and $c=\lVert x \rVert ^2$. This is a rel-valued quadratic polynomial in the variable $t$. Since this polynomial is nonnegative, $\color{red}{it\;can\;have\;at\;most\;\;one\;real\;root}.$
$\color{blue}{This\;implies\;that\;the\;discriminant\; cannot\;be\;strictly\;positive}.$

Question. What are the reasons why the red and blue assertions hold? I need the precise details.

Thanks!
 A: Take example of $x^2-3x+2 = (x-1)(x-2) = 0$. It has two roots and between $x = 1$ and $x = 2$, it is negative, while positive for $x \lt 1, x \gt 2$. As the quadratic in the question is always non-negative (zero or positive), it does not change sign so either it may have no root or at most it may have one root (which is both roots having the same value e.g. $x^2 - 2x + 1 = (x-1)(x-1) = 0$
When the quadratic $(ax^2 + bx + c)$ has two real roots, its discriminant is positive i.e $b^2 - 4ac \gt 0$ and if it has only one real root its discriminant is zero i.e $b^2-4ac = 0$. You can check the same for above two quadratic examples in my answer. If the discriminant is negative i.e. $b^2 - 4ac \lt 0$, it will not have any real roots.
A: The statements are related to the quadratic equation
$$ax^2+bx+c=0 \implies x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$$
since when $b^2-4ac>0$ then $\sqrt{b^2-4ac}$ exists and we always have two distinct real solutions.
Moreover for the quadratic function
$$f(x)=ax^2+bx+c$$
which represents a parabola, when $f(x)=0$ has two solutions then $f(x)$ takes positive and negative values.
A: There's a high-school  theorem on the sign of a quadratic polynomial:

A quadratic polynomial $p(x)=ax^2+bx+c\quad(a\ne 0)$ has the sign of its leading coefficient, except between its (real) roots, if any.

