I'm thoroughly confused by the Schwarz inequality.(adsbygoogle = window.adsbygoogle || []).push({});

In many books on analysis, the author defines the Euclidean distance and then shows that the distance function is a metric, ie the Schwarz inequality holds. This isn't very satisfying as I feel it masks the properties of R^n that generate this relation.

In reading on Wikipedia, I came across the topological idea of a uniform space. Although I don't quite understand the characterization of a uniform space, it satisfied me to read that we could salvage some parts of analysis without resorting to a real-valued distance function. We can still formalize the notion of "relative closeness".

From a simple algebraic viewpoint, one can easily define an absolute value in an ordered ring. This order relation satisfies the triangle inequality, and I also believe it satisfies the Schwarz inequality. So is every ordered ring a "metric space" where the standard metric takes on values from the ring?

This makes sense until we look at the complex plane. It's not an ordered field (or ring) but we can still define a absolute value. This work because we identify the complex plane with R^2... but that brings me back the original question, what properties of R^n allow us to prove the Schwarz inequality?

If this seems like a half-baked post, it is... I'm having a hard time even formulating my question and wanted to see if somebody could direct me to further reading on these ideas (my home library is coming up short).

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# Schwarz Inequality

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