The general intution as to why the vector with the norm of 0 is chosen during the Cauchy-Schwarz inequality proof. My professor mentioned that reason $t$ was chosen during the Cauchy-Schwarz inequality proof can also be seen when we minimize $\lVert x\rVert^2 +2t\langle x,y \rangle +t^2\lVert y\rVert^2$ over $t \in \mathbb{R}$. I'm not really sure what he is getting at.
 A: We know that 
$$
|| x + ty||^2 \geq 0
$$
for all $t \in \mathbb{R}$.
Then we have
$$
0 \leq ||x+ty||^2 = ||x||^2 + 2t\langle x, y\rangle + t^2||y||^2.
$$
Hence the inequality
$$
||x||^2 + 2t\langle x, y\rangle + t^2||y||^2 \geq 0
$$
holds for all $t \in \mathbb{R}$. Now what is the "best" $t$ that we could choose?
The smaller the left hand side is, the tighter a bound we have proven. This gives us the idea to minimize the left side in $t$. One may differentiate to find
$$
2\langle x, y \rangle + 2t ||y||^2 = 0 \implies t = \frac{-\langle x ,y \rangle}{||y||^2}
$$
is where the minimum occurs (check that it is indeed a minimum using any method you prefer). If $||y||=0$ then $y=0$ and the Cauchy-Shwartz inequality holds anyway, so we exclude this case.
Plugging in that value of $t$ and rearranging then gives $\langle x, y \rangle^2 \leq ||x||^2 ||y||^2$ and taking square roots gives the Cauchy-Shwartz inequality.
Note that a much simpler way to derive this is to note that
$$
0 \leq ||x+ty||^2 = ||x||^2 + 2t\langle x, y\rangle + t^2||y||^2
$$
says that a parabola has at most one real root, so its disciminant $b^2-4ac$ must be nonpositive, i.e. 
$$
(2\langle x, y\rangle)^2 - 4 (||x||^2) (||y||^2) \leq 0
$$
which givees the desired inequality immediately.
