How does this proof of the Cauchy-Schwarz Inequality work? I was watching a Khan Academy video on the Cauchy-Schwarz Inequality, and I just can't seem to understand the proof, and the comments on the video don't seem to help.  The video is here.
First, he creates an artificial function:
$$
p(t)=||t\vec{y}-\vec{x}||^2\geq0
$$
What is the motivation behind this function?  What does it mean?
Next, after substituting in the dot product of $t\vec{y}-\vec{x}$ with itself, he obtains:
$$
p(t)=(\vec{y}\cdot\vec{y})t^2-2(\vec{x}\cdot\vec{y})t+\vec{x}\cdot\vec{x} \geq 0\\a\equiv(\vec{y}\cdot\vec{y})\\b\equiv2(\vec{x}\cdot\vec{y})\\c\equiv\vec{x}\cdot\vec{x}\\p(t)=at^2-bt+c \geq 0
$$
I get this part, but it's the next part that confuses me; he chooses $t=\frac{b}{2a}$.  Why choose that particular value for t?  At first I thought that it had something to do with minimizing the value of the function, but then I realized that it's not $-\frac{b}{2a}$; which further confuses me as for the significance of the chosen value.
I'm currently a high school student taking AP Calculus trying to study linear algebra on my own, so if it's not too much trouble, please dumb things down for me a bit.
 A: Sal's proof here works by constructing this function $p(t)$, showing that $p(t)\geq 0$ everywhere, and then using that fact to bound the coefficients of $p$, which thankfully yields the Cauchy-Schwarz inequality. So once he establishes that $p(t)\geq 0$ for all $t$ and shows that $p(t)=at^2-bt+c$ for appropriate choices of $a, b,$ and $c$, the next step is to use this to get information about the coefficients $a,b,c$. Now, it is clear that different choices of $t$ will give us different information about the coefficients—choosing $t=10^{50}$ gives us
$$(\mathbf{y}\cdot\mathbf{y})10^{100}-2(\mathbf{x}\cdot\mathbf{y})10^{50}+(\mathbf{x}\cdot\mathbf{x})\geq 0$$
which is certainly true, but fairly obvious and hardly helpful. In order to exploit the $p(t)\geq 0$ inequality in a useful way, it makes sense to choose $t$ which makes $p(t)$ as small as possible; this will give us the sharpest bound and thus, in a sense, the most information about the coefficients. And information about the coefficients is what we are looking for. So we choose $t$ which minimizes $p(t)$ which is, of course, $b/(2a)$.
I prefer a slightly different approach which uses the same function and essentially works the same. Define this same function $p(t)$, so we get $$
\begin{align}
p(t) &= \left\Vert t\mathbf{y}-\mathbf{x}\right\Vert^2 \\
&=(t\mathbf{y}-\mathbf{x})\cdot (t\mathbf{y}-\mathbf{x})\\
&=(\mathbf{y}\cdot\mathbf{y})t^2-2(\mathbf{x}\cdot\mathbf{y})t+(\mathbf{x}\cdot\mathbf{x}) \\&\geq 0
\end{align}
$$
This is a quadratic polynomial in $t$ which has at most one real root (can you see why?). As you'll recall from your grade 11 algebra class, a quadratic has at most one real root if the discriminant is 0 or negative, so we obtain
$\Delta=4(\mathbf{x}\cdot\mathbf{y})^2-4(\mathbf{y}\cdot\mathbf{y})(\mathbf{x}\cdot\mathbf{x})\leq0$
