Why does the Cauchy-Schwarz Inequality even have a name?
Side note: it's actually the Cauchy-Schwarz-Bunyakovsky inequality, and don't let anyone tell you otherwise.
The problem with using the geometric definition is that you have to define what an angle is. Sure, in three dimensional space, you have pretty clear ideas about what an angle is, but what do you take as $\theta$ in your equation when $i$ and $j$ are $10$ dimensional vectors? Or infinitely-dimensional vectors? What if $i$ and $j$ are polynomials?
The Cauchy-Schwarz inequality tells you that anytime you have a vector space and an inner product defined on it, you can be sure that for any two vectors $u,v$ in your space, it is true that $\left|\langle u,v\rangle\right| \leq \|u\|\|v\|$.
Not all vector spaces are simple $\mathbb R^n$ businesses, either. You have the vector space of all continuous functions on $[0,1]$, for example. You can define the inner product as
$$\langle f,g\rangle=\int_0^1 f(x)g(x)dx$$
and use Cauchy-Schwarz to prove that for any pair $f,g$, you have $$\left|\int_{0}^1f(x)g(x)dx\right| \leq \sqrt{\int_0^1 f^2(x)dx\int_0^1g^2(x)dx}$$
which is not a trivial inequality.
The inequality is ubiquitous, so some name is needed.
As there is no cosine in the statement of the inequality, it cannot be called "cosine inequality" or anything like that.
The geometric interpretation with cosines only works for finite-dimensional real Euclidean space, but the inequality holds and is used more generally than that. That is Schwarz' contribution.
Schwarz founded the field of functional analysis (infinite-dimensional metrized linear algebra) with his proof of the inequality. That is important enough to warrant a name. In terms of consequences per line of proof it is one of the greatest arguments of all time.
The Schwarz proof was part of the historical realization that Euclidean geometry, with its mysterious angle measure that seems to depend on notions of arc-length from calculus, is the theory of a vector space equipped with a quadratic form. That is a major shift in viewpoint.
Stating the inequality in terms of cosines assumes that the inner product restricts to the standard Euclidean one on the 2 (or fewer) dimensional subspace spanned by the two vectors, and that you have proved the inequality for holds for standard Euclidean space of 2 dimensions or less. How do you know those things are correct without a much longer argument? That argument will, probably, include somewhere a proof of the Cauchy-Schwarz inequality, maybe written for 2-dimensions but working for the whole $n$-dimensional space, so it might as well be stated as a direct proof for $n$ dimensions. Which is what Cauchy and Schwarz did.
Cauchy-Schwarz is not just that. The result that you stated is just a special case of Cauchy-Schwarz in Euclidean spaces. But it's still valid in any inner product space, equipped with any inner product. The proof is still easy though, but nobody said that the proof had to be long and difficult to give it a name. The fact is that Cauchy-Schwarz inequality is very useful in many applications, from geometry to probability theory, and that's why it's worth having its own name.