In Rudin's Theorem 1.37(d), he says that $|v \cdot w| \leq |v||w|$ for vectors $v, w \in \mathbb{R^n}$. He claims that this is "an immediate consequence of the Schwarz inequality."
I cannot tell what he means. In my understanding, this is the Cauchy-Schwarz inequality. What is ruding stating that this follows from? I know that he proved a more gneeral version of Cauchy-Schwarz using complex numbers. Is he using that in some way to prove this?
Well, think about what the dot product of two vectors in $\mathbb{R^n}$ means and what their norm refers to.
If we let $v = (v_1,v_2,\ldots,v_n)$ and $w = (w_1,w_2,\ldots,w_n)$, then:
$$|v \cdot w| = |v_1 \cdot w_1 + \ldots + v_n \cdot w_n|$$
$$|v| \cdot |w| = \sqrt{v_1^2+v_2^2 + \ldots v_n^2} \cdot \sqrt{w_1^2+w_2^2+\ldots+w_n^2}$$
So, if you've proved the Cauchy-Schwarz Inequality, you can show that the formula Rudin gives follows from that just by using the definition of the standard inner product on $\mathbb{R^n}$.
In other words, the Cauchy-Schwarz Inequality for real numbers says nothing about vectors or vector spaces. By giving $\mathbb{R^n}$ a vector space structure and a dot product, you can write the inequality in a more concise manner that looks beautiful.
Anyways, I don't particularly like this order that Rudin has taken with proving that inequality. I prefer the more general approach where we show that the Cauchy-Schwarz Inequality holds for any Euclidean Vector Space using the properties of a general inner product.
Once you have done that and recognize that the standard inner product makes $\mathbb{R^n}$ a Euclidean Vector Space, the Cauchy-Schwarz Inequality is applicable there and the standard formulation of it immediately follows. I feel there's a bit more beauty in doing it that way.