Cauchy-Schwarz Inequality : What motivates, and how can you divine, the quadratic polynomial?

Question

Source: Poole, D. Linear Algebra: A Modern Introduction (2014 4 edn). Section 7.1. Exercise 44.

44. Let $\mathbf{u}$ and $\mathbf{v}$ be vectors in an inner product space $V$. Prove the Cauchy-Schwarz Inequality for $\mathbf{u\neq0}$ as follows: (a) Let $t$ be a real scalar. Then $\langle t\mathbf{u} + \mathbf{v}, t\mathbf{u} + \mathbf{v}\rangle \ge 0$ for all values of $t$. Expand this inequality to obtain a quadratic inequality of the form $at^2 + bt + c \ge 0$ [...]

How can you divine this trick of inventing this quadratic polynomial? It feels like a flash of genius, but I'm not clairvoyant. I already know, and ask not, how to execute the algebra.

Timothy Gowers feels likewise that

most textbooks and all analysis courses I have attended favour the approach where you write down

$ \lVert x-cy \rVert^2$, which is real and non-negative, and then choosing a `clever' value of $c$ from which to deduce the Cauchy-Schwarz inequality. Of course, $c$ can be justified as the value that minimizes the quadratic expression that results from expanding $\lVert x-cy \rVert^2$, but even so the idea of writing down $ \lVert x-cy \rVert^2$ in the first place is not an obvious one [mine].

No explanation is usually given of where the quadratic form comes from. This page is intended for those who happen not to have observed, or been shown, that more or less the same argument can be made to seem much more natural. Indeed, this is another example of a proof that a well-programmed computer could reasonably be expected to discover.

Answer 1

For ${\Bbb C}^n$ one may write $$ \|x\|^2\|y\|^2-|\langle x,y\rangle|^2\ge 0\quad\Leftrightarrow\quad\begin{bmatrix}x^*x & x^*y\\y^*x & y^*y\end{bmatrix}=\begin{bmatrix}x \\y\end{bmatrix}^*\begin{bmatrix}x \\y\end{bmatrix} \text{ pos.semidef.} $$ and the latter is obvious for all $x$, $y$. How would one test positive semidefiniteness of the Gramian matrix $$ \begin{bmatrix}\langle x,x\rangle & \langle x,y\rangle\\\langle y,x\rangle & \langle y,y\rangle\end{bmatrix} $$ in a general Hilbert space? To pre- and postmultiply by the vector $(1,t)$ sound like a natural way to do it.

Answer 2

I think the way to derive it systematicaly could be something like the following:

1. We have an inequality that includes quadratics $$||\vec{u} \cdot \vec{v}||^2 \le ||\vec{u}||^2 ||\vec{v}||^2$$

thus we need some quadratic inequality to begin with.

2. A qudratic inequality which holds by defintion in an inner product space is the $||\vec{u}||^2 \ge 0$

3. Use a linear combination of the two vectors and apply inequality in (2) to try to derive the needed inequality. One such combination is to remove the projection of the first vector onto the second. This has a geometric meaning and applies to the special case where the inequality equals zero (linearly dependent). ie $0 \le ||\vec{u}-\lambda\vec{v}||$