Proof of Cauchy-Schwarz inequality.

Question

I'm trying to understand the proof of the Cauchy-Schwarz inequality: for two elements x and y of an inner product space we have

$$\lvert \langle x,y\rangle\rvert \leq\lVert x \rVert \cdot\lVert y\lVert$$

The proof I am reading goes as follows:

We may assume that $y\neq0$ and $\lVert y\rVert=1$. Indeed, the Cauchy-Schwarz inequality holds when y=0. If $y\neq0$ then $z=\frac{y}{\lVert y\rVert}$ has length 1. So if $\lvert \langle x,z\rangle\rvert\leq \lVert x\rVert$ holds then $$\lvert \langle x,z\rangle\rvert=\frac{\langle x,y\rangle}{\lVert y\rVert}\leq\lVert x\rVert$$ from which $\lvert \langle x,y\rangle\rvert \leq\lVert x\rVert \cdot\lVert y\rVert$ follows.

The confusing part is in bold, why does this inequality hold in general?

Answer 1

The given argument in the question is not much of a proof, since the "hard" part is to prove that for unit $z$ we have $|<x,z>| \le ||x||$. It certainly admits the interpretation in terms of projection given by cheepychappy, however, this is nothing less than the interpretation of the Cauchy-Schwarz inequality itself.

The proof i recommend is as follows. Consider $x,y$ any vectors in a complex vector space. Then for any $r>0$ and any real $\theta$ expand the inequality $<x+re^{i\theta}y,x+re^{i\theta}y> \ge 0$ to obtain a condition on the roots of a polynomial of second degree in the variable $r$. Express this condition in terms of the coefficients of the polynomial (which will include $||x||, ||y||, |<x,y>|$). What you will get will be the C-S inequality.

Answer 2

If you think of the inner product $\lvert \langle x,z\rangle\rvert$ as being the projection of $x$ in the direction of $z$, and that projection has the length $\lVert x \rVert \cdot \lVert z \rVert \cdot \cos(\theta)$, the fact that $\cos(\theta)$ has an upper-bound of $1$, means that $\lvert \langle x,z\rangle\rvert\leq \lVert x\rVert$ whenever $\lVert z \rVert = 1$.