Proof of the Cauchy-Schwarz inequality in Axler's Linear Algebra Done Right The proof of the Cauchy-Schwarz inequality in Axler's Linear Algebra Done Right (pg. 104) hinges on showing that
$\|u\|^2 = \|\frac{\langle u,v\rangle}{||v||^2}v\|^2 + \|w\|^2 \tag{1}$
equals $\frac{|\langle u,v\rangle|^2}{\|v\|^2} + \|w\|^2 \tag{2}$
but I'm not quite sure how (1) = (2). 
Focusing in on $\|\frac{\langle u,v\rangle}{\|v\|^2}v\|^2$ we can rewrite this as:
$\|\frac{\langle u,v\rangle}{\langle v,v\rangle}v\|^2$ where the fraction is a scalar so we can take it out: $|\frac{\langle u,v\rangle}{\|v\|^2}|^2\|v\|^2$ and then the squared norms of $v$ cancel out, but clearly this wrong somehow.
Many thanks for any help.
 A: You've solved the question, so I can't understand your confusion. See this:
$$\|\frac{\langle u,v\rangle}{\|v\|^2}v\|^2=|\frac{\langle u,v\rangle}{\|v\|^2}\times \|v\||^2=\frac{|\langle u,v\rangle|^2}{\|v\|^2}$$
Hope this can help you.
A: For some reason, people teaching Linear Algebra forgot to connect inner product with the ordinary dot product in $\mathbb{R}^{3}$ or $\mathbb{R}^{n}$. If you have a vector $u$ and a line through the origin with direction vector $v$, then how would you find the point on that line closest to $u$? Answer: orthogonal projection. The same holds in infinite dimensional inner-product spaces, and even for complex spaces. You want to find a scalar $\alpha$ such that $(u-\alpha v)\perp v$, which gives
$$
                         (u,v)-\alpha(v,v) = 0,\\
                         \alpha = \frac{(u,v)}{(v,v)}
$$
Now you can decompose into a right triangle where $u$ is the hypotenuse, $\alpha v$ is the leg along the line with direction vector $v$ and $u-\alpha v$ is the other leg. Explicitly, you have the following orthogonal decomposition:
$$
            u = \alpha v + (u-\alpha v),\\
              (\alpha v,u-\alpha v) = 0.
$$
By the Pythagorean Theorem (which is a direct computation using inner product axioms):
$$
                  \|u\|^{2} =\|\alpha v\|^{2}+\|(u-\alpha v)\|^{2}.
$$
The Cauchy-Schwarz inequality is exactly the following
$$
                       \|\alpha v\|^{2} \le \|u\|^{2} \\
            \mbox{ with equality iff } \|u-\alpha v\| = 0.
$$
Write this out:
$$
                       |\alpha|^{2}\|v\|^{2} \le \|u\|^{2} \\
                        \frac{|(u,v)|^{2}}{(v,v)^{2}}\|v\|^{2} \le (u,u) \\
                        |(u,v)|^{2} \le (u,u)(v,v).
$$
