TITLE: Why does the Cauchy-Schwarz inequality hold in any inner product space?
QUESTION [9 upvotes]: I am working through linear algebra problems in Apostol's Calculus, and he has numerous problems that seem to imply that Cauchy-Schwarz holds no matter how the inner product is defined. Then, he has problems where the triangle inequality holds despite alternative definitions of vector norm. This got me thinking, since the proof of the triangle inequality in Apostol relies on Cauchy-Schwarz, that the triangle inequality would hold regardless of how the vector norm is defined (if it involves the dot product).
I then found this response to a question, which states what I was thinking.
Are there any proofs that the Cauchy-Schwarz inequality holds in any inner product space (I looked for some and found none and could not prove it myself)? I've had a semester of algebra (Artin) and some analysis, if there is a proof at such a level of understanding. Intuitive explanations are good, too.

REPLY [2 votes]: Here is an alternative perspective: Cauchy-Schwarz inequality holds in every inner product space because it holds in $\mathbb C^2$. On p.34 of Lectures on Linear Algebra, Gelfand wrote:

Any 'geometric' assertions pertaining to two or three vectors is true if it is true in elementary geometry of three-space. Indeed, the vectors in question span a subspace of dimension at most three. This subspace is isomorphic to ordinary three space (or a subspace of it), and it therefore suffices to verify the assertion in the latter space. In particular the Schwarz inequality -- a geometric theorem about a pair of vectors -- is true because it is true in elementary geometry.

In our case, Gelfand's remark translates to the following. Let $x$ and $y$ be any two vectors in an inner product space $V$ over a field $\mathbb F=\mathbb R$ or $\mathbb C$. Let $S_V=\operatorname{span}\{x,y\}$ and $m=\dim S_V$. By defining an $\mathbb F$-linear map $T$ that maps an orthonormal basis of $S_V$ to an orthonormal basis of a subspace $S_{\mathbb C^2}$ of $\mathbb C^2$ of dimension $m$, one may verify that $T$ is an isometry, i.e. $\langle u,v\rangle_V=\langle Tu,Tv\rangle_{\mathbb C^2}$ for any $u,v\in S_V$. It follows that $|\langle x,y\rangle_V|\le\|x\|_V\|y\|_V$ if and only if $|\langle Tx,Ty\rangle_{\mathbb C^2}|\le\|Tx\|_{\mathbb C^2}\|Ty\|_{\mathbb C^2}$, but the latter is just Cauchy-Schwarz inequality in $\mathbb C^2$.
So, if you can prove that Cauchy-Schwarz inequality holds in $\mathbb C^2$, it must also hold in every inner product space $V$ over $\mathbb R$ or $\mathbb C$, even when $V$ is infinite-dimensional (but $\mathbb C^2$ is not).
Note the subtle difference between this answer and the others. In each of the other answers, a generic proof is given such that the inequality is proved once and for all for every inner product space. Here, we don't prove the inequality directly. Instead, we show that the inequality in a general inner product space follows from its specific version $\mathbb C^2$. How it is actually proved in $\mathbb C^2$ is not our concern. Of course, in order to prove it in $\mathbb C^2$, we may end up using the proofs given in the other answers. However, the point Gelfand made is that if we have learnt a proof of Cauchy-Schwarz inequality in $\mathbb C^2$ in the past, then we don't need to worry whether the proof is generic or not, because the validity of the inequality is automatically generic.