Editing Cauchy–Schwarz inequality (section)

===Proof by analyzing a quadratic===

Consider an arbitrary pair of vectors <math>\mathbf{u}, \mathbf{v}</math>.  Define the function <math>p : \R \to \R</math> defined by <math>p(t) = \langle t\alpha\mathbf{u} + \mathbf{v}, t\alpha\mathbf{u} + \mathbf{v}\rangle</math>, where <math>\alpha</math> is a complex number satisfying <math>|\alpha| = 1</math> and <math>\alpha\langle\mathbf{u}, \mathbf{v}\rangle = |\langle\mathbf{u}, \mathbf{v}\rangle|</math>.
Such an <math>\alpha</math> exists since if <math>\langle\mathbf{u}, \mathbf{v}\rangle = 0</math> then <math>\alpha</math> can be taken to be 1.

Since the inner product is positive-definite, <math>p(t)</math> only takes non-negative real values. On the other hand, <math>p(t)</math> can be expanded using the bilinearity of the inner product:
<math display=block>\begin{align}
p(t)
&= \langle t\alpha\mathbf{u}, t\alpha\mathbf{u}\rangle + \langle t\alpha\mathbf{u}, \mathbf{v}\rangle + \langle\mathbf{v}, t\alpha\mathbf{u}\rangle + \langle\mathbf{v}, \mathbf{v}\rangle \\
&= t\alpha t\overline{\alpha}\langle\mathbf{u}, \mathbf{u}\rangle + t\alpha\langle\mathbf{u}, \mathbf{v}\rangle + t\overline{\alpha}\langle \mathbf{v}, \mathbf{u}\rangle + \langle\mathbf{v}, \mathbf{v}\rangle \\
&= \lVert \mathbf{u} \rVert^2 t^2 + 2|\langle\mathbf{u}, \mathbf{v}\rangle|t + \lVert \mathbf{v} \rVert^2
\end{align}</math>
Thus, <math>p</math> is a polynomial of degree <math>2</math> (unless <math>\mathbf{u} = 0,</math> which is a case that was checked earlier). Since the sign of <math>p</math> does not change, the discriminant of this polynomial must be non-positive:
<math display=block>\Delta = 4 \bigl(\,|\langle \mathbf{u}, \mathbf{v} \rangle|^2 - \Vert \mathbf{u} \Vert^2 \Vert \mathbf{v} \Vert^2\bigr) \leq 0.</math>
The conclusion follows.<ref>{{Cite book|title=Real and Complex Analysis|last=Rudin|first=Walter|publisher=McGraw-Hill|year=1987|isbn=0070542341|edition=3rd|location=New York|orig-year=1966}}</ref>

For the equality case, notice that <math>\Delta = 0</math> happens if and only if <math>p(t) = \bigl(t\Vert \mathbf{u} \Vert + \Vert \mathbf{v} \Vert\bigr)^2.</math> If <math>t_0 = -\Vert \mathbf{v} \Vert / \Vert \mathbf{u} \Vert,</math> then <math>p(t_0) = \langle t_0\alpha\mathbf{u} + \mathbf{v},t_0\alpha\mathbf{u} + \mathbf{v}\rangle = 0,</math> and hence <math>\mathbf{v} = -t_0\alpha\mathbf{u}.</math>