The Wikipedia article inner product space
gives the properties that an inner product has to satisfy.
If you already know and understand these properties then
you can skip to the summary at the end of this answer.
The article doesn't mention why we have inner products.
Suppose we have a finite dimensional real vector space such
as $\,\mathbb{R}^n\,$. In this form, each vector is a tuple
$\,(a_1,a_2,\dots,a_n)\,$ real numbers. However, in general,
although a vector space $\,V\,$ is assumed to have a finite
basis, how to find the coefficients $\,a_k\,$ of any vector?
The solution is to use the dual space $\,V^*\,$ with its corresponding dual basis.
The Wikipedia article states
The pairing of a functional $\,\varphi\,$ in the dual space
$\,V^∗\,$ and an element $\,x\,$ of $\,V\,$ is sometimes denoted
by a bracket: $\,\varphi(x) =[x,\varphi]\,$ or
$\,\varphi(x)=\langle\varphi,x\rangle.\,$
This pairing defines a nondegenerate bilinear mapping
$\,\langle,\cdot,\cdot\rangle: V^* \times V
\to F\,$ called the natural pairing.
Later in the article it states:
If $\,V\,$ is finite-dimensional, then $\,V\,$ is isomorphic
to $\,V^*\,$. But there is in general no natural isomorphism
between these two spaces. Any bilinear form
$\,\langle\cdot,\cdot\rangle\,$ on $\,V\,$ gives a mapping
of $\,V\,$ into its dual space via
$\,v\mapsto\langle v,\cdot\rangle\,$ where the right hand
siade is defined as the functional on $\,V\,$ taking each
$\,w\in V\,$ to $\,\langle v,w\rangle\,$.
Given a nondegenerate inner product we choose an rthonormal
basis $\,\{e_i\}\,$ and we can now write any vector
$\,v\in V\,$ as
$\,v = a_1\,e_1+a_2\,e_2+\cdots+a_n\,e_n\,$. where the
coefficients $\,a_i\,$ are the equivalents to the elements of
$n$-tuples in $\,\mathbb{R}\,$.
With this background, consider any vector $\,v\in V\,$. Without
loss of generality, choose any unit vector $\,w\,$. We can write
$$v = \langle v,w\rangle\, w + t \tag{1} $$
where $\,t\,$ consists of
the sum of the remaining components of $\,v\,$. We find that
$\;t\!\perp\!w\;$ because from equation $(1)$ we get that
$$\langle v,w\rangle = \langle v,w\rangle \langle w,w
\rangle+ \langle t,w\rangle\tag{2}$$
and since $\,\langle w,w\rangle=1\,$ by assumption on $\,w\,$, implies that $\,\langle t,w\rangle=0\,$. Also
from equation $(1)$ we get that
$\,\langle v,t\rangle = \langle v,w\rangle \langle w,t\rangle
+ \langle t,t\rangle\,$. But we showed that
$\,\langle t,w\rangle=0\,$ which implies that
$\,\langle v,t\rangle = \langle t,t\rangle\,$ and this implies
$\,\langle v-t,w\rangle=0\,$ or $\;(v-t)\!\perp\!w\;$.
This proves that $\,0,v\,$ and $\,t\,$ form the vertices
of a right triangle with the right angle at $\,t\,$.
The Pythagorean theorem in this case states that
$$ ||v-t||^2 + ||0-t||^2 = ||0-v||^2. \tag{3}$$
By equation $(1)$ we get
$\,v-t = \langle v,w\rangle\, w\,$ which by
assumption on $\,w\,$ implies
$\,||v-t||^2 = \langle v,w\rangle^2\,$ and from equation
$(3)$ we get that $\, \langle v,w\rangle^2\le ||v||^2\,$.
Without the assumption that $\,\langle w,w\rangle=1\,$
we get the general inequality
$$ \langle v,w\rangle^2 \le ||v||^2\,||w||^2. \tag{4}$$
Summary: The visual interpretation of the Cauchy
Schwarz inequality and its standard proof is that
the vector $\,v\,$ is decomposed into a vector
$\,t\,$ perpendicular to $\,w\,$ and the vector
$\,v-t\,$. These three vectors form a right
triangle. The Pythagorean theorem implies the
Cauchy Schwarz inequality since one consequence
of the theorem is that in any right triangle the
length of any leg is bounded by the length of the
hypoteneuse.
