I was working on an algebraic proof for the Triangle Inequality when I came across an instance in which $$(a \cdot b) \le ||a|| \cdot ||b||$$ Is this true in general, or was this just a special case? Thanks in advance!
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Search the net (or MSE) for Cauchy-Schwarz inequality. – Kavi Rama Murthy Jan 12 '22 at 05:08
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@KaviRamaMurthy Right, of course! I somehow completely forgot that, despite it being the reason for me working on this in the first place lol thank you! – bjorn Jan 12 '22 at 05:12
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It is the Cauchy-Schwarz (CS) ineq. as Kavi points out. Since you mention proving the triangle ineq., I will also point out a few things:
CS holds in any inner product space.
The triangle ineq. holds for any normed vector space; in fact, it is a defining feature of a norm.
Using CS to show the triangle ineq. actually explains why any inner product $\langle\cdot,\cdot\rangle$ induces a valid norm defined by
$$\|x\|\equiv \sqrt{\langle x,x \rangle}.$$
Indeed, any inner product space is always a normed space with the norm thus defined, but the converse is not true. There are norms not induced by inner products (e.g. any $p$ norm, $p\neq 2$). A proof of the triangle inequality using CS would thus not apply for such cases.
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