Proof of Cauchy-Schwarz in $\mathbb{R}^n$ using Law of Cosines Let us consider the Cauchy-Schwarz Inequality for vectors in $\mathbb{R}^n$:
Let $u, v \in \mathbb{R}^n$. Then $|u \cdot v| \leq |u||v|$.
Wikipedia provides a proof for this setting by enforcing a condition on the discriminant of a polynomial.
What I don't understand is why this specific case of Cauchy-Schwarz doesn't immediately follow from the law of cosines $u \cdot v = |u| |v| \cos{\alpha}$, where $\alpha$ is the angle between $u$ and $v$.
We know that the span of $u$ and $v$ is at most two-dimensional, and so if we restrict ourselves to $\mathrm{Span}(u,v)$ we are in standard Euclidean geometry, the angle $\alpha$ is well-defined, and the law of cosines holds.
I would be very grateful if someone could point out an error in the argument I've made (a quick proof of Cauchy-Schwarz for vectors in $\mathbb{R}^n$ as an immediate consequence of the law of cosines, without any recourse to discriminants of polynomial equations); or, if I haven't made an error, a perspective on why Wikipedia and other references (like Hubbard and Hubbard, Vector Calculus, Theorem 1.4.5) prefer what seems to me a more complicated and less intuitive proof.
 A: In the Euclidean space we can prove that $u \cdot v = |u| |v| \cos{\alpha}$ holds which is equivalent to Cauchy-Schwarz inequality and $\cos{\alpha}$ corresponds to the usual definition of cosine.
Cauchy-Schwarz inequality holds in general for any vector space with dot product and by this we can extend the definition of $\cos {\alpha}$ which is given by
$$ \cos{\alpha}=\frac{u \cdot v}{  |u| |v|}$$
A: Your reduction to two dimensions is valid, but in practice it requires you to also show $u\cdot v=|u||v|\cos\alpha$.holds in $2$ dimensions, and that $\cos\alpha\in[-1,\,1]$ for any angle in the plane, i.e. $\alpha\in\Bbb R$ (give or take your favourite modulo-$2\pi$ restriction convention). I suspect you already know how to prove these results, but they do take a little work.
I think Wikipedia just wanted to show you there's a purely non-trigonometric way to do it, which isn't surprising given that CS can be stated as an inequality in real variables, no matter how little the reader knows about vector spaces.
It's also worth noting that algebraic methods, while often written with a real vector space in mind, are very easily tweaked to deal with complex vector spaces instead. Is it obvious one can define an angle, equal to a real number of radians, between two vectors spanning a complex plane?
