Intuition for the Cauchy-Schwarz inequality I'm not looking for a mathematical proof; I'm looking for a visual one. I'm having trouble understanding (in my mind's eye) why the dot product of two vectors V and W produces a scalar that is less than the length of V multiplied by the length of W.
In using the dot product, we are producing a parallel vector, correct? Could we not further say that we are simply applying vector W to vector V in order to produce a vector that is the original length of V multiplied by the length of W -- thus a vector parallel to V? For example, if we let vector W be a unit vector (with length of one), then the dot product of V and W would give us a scalar that, when applied to V, produces V again. Would this not be the same as the length of V multiplied by the length of W (given that the length of W is equal to one)?
For that reason, why wouldn't the dot product of V and W always be equal to the length of V multiplied by the length of W? Why would it be less (unless V = cW for any scalar c?)
 A: By definition, the "dot" product of two vectors, say $\vec A$ and $\vec B$ is
$$\vec A\cdot \vec B=|\vec A||\vec B|\cos \theta$$
where $\theta$ is the angle between $\vec A$ and $\vec B$.  That is to say, that the inner product is the projection of one vector onto the other.  Visually, the projection is like a "shadow"  that one vector casts along the direction of the other.
A: In the Cauchy–Schwarz (CS) inequality $|u\cdot v|\le \|u\|\|v\|$, let's assume $v$ is a normalised vector, i.e., $\|v\|=1$. Then the CS inequality becomes $|u\cdot v|\le \|u\|$. Now, it's a trivial matter to show that these two forms of the CS inequality are in fact equivalent, in the sense that if $|u\cdot v|\le \|u\|$ for all normalised vectors $v$, then the usual CS inequality holds for all vectors. So, let us restate the CS inequality as stating that $|u\cdot v|\le \|u\|$ for all normalised vectors $v$. Now, the physical/geometric interpretation of $u\cdot v$ in this case is that it is the component of the vector $u$ in the direction $v$ (since $v$ is assumed normalised, that's all it is, a direction), while $\|u\|$ is the magnitude of $u$. So the CS inequality is merely stating the intuitively obvious fact that the component of a vector $u$ in a single direction is bounded by the magnitude of $u$.  
Incidentally, this line of thought carries on to produce a very short and elegant proof of the full CS inequality. But, as you are not looking for a proof, I'll leave that out as an exercise. 
A: One can show that in Euclidean space, the angle $\theta$ between two vectors $v,w$ (in the sense of Euclidean geometry) satisfies
$$\cos(\theta)=\frac{v \cdot w}{\| v \| \| w \|}.$$
This is basically the law of cosines applied to an appropriate triangle. This equation only makes sense for every $v,w$ if the Cauchy-Schwarz inequality holds.
A: Recall that $$a\cdot b=|a||b|\cos\theta$$ where $\theta$ is the angle between $a$ and $b$.  
Using this fact it is easy to check that $\dfrac{a\cdot b}{|b|}$ is the component of $a$ in the direction of $b$.  Of course the component of $a$ in the direction of $b$ must have absolute value less than or equal to the magnitude of $a$.  This gives $\dfrac{|a\cdot b|}{|b|}\leq|a|$ and hence $|a\cdot b|\leq |a||b|$.  
So really $a\cdot b=|a||b|\cos\theta$ gives not only a formal proof of the Cauchy-Schwarz inequality, but also a geometric way of thinking of the dot product that makes the Cauchy-Schwarz inequality clear.  
A: 
(Adapted from wikimedia commons: File:Dot Product.svg using Inkscape 0.91 to convert to PNG.)
The image illustrates the scalar projection of $\mathbf{A}$ onto $\mathbf{B}$, sometimes denoted $A_B$.  You already know that, if $||\mathbf{B}||=1$, $\mathbf{A} \cdot \mathbf{B} = A_B$, and so for nonspecial $\mathbf{B}$,
$$ \mathbf{A} \cdot \mathbf{B} = \mathbf{A} \cdot \hat{\mathbf{B}}||\mathbf{B}|| =  A_B ||\mathbf{B}|| = ||\mathbf{A}|| \, ||\mathbf{B}|| \cos \theta$$
where $\hat{\mathbf{B}}$ denotes the unit vector along $\mathbf{B}$.
But what does this tell us?  That $\mathbf{A} \cdot \mathbf{B}$ is maximized when $\theta$ is 90 degrees.  In that case, the parallelogram $\mathbf{0}, \mathbf{A} , \mathbf{A} + \mathbf{B}, \mathbf{A} + \mathbf{B} - \mathbf{A} ({}=\mathbf{B})$ is a rectangle.  Using the area formula for parallelograms (base times height), the area is maximized when $\mathbf{A}$ is all height.  When $\theta$ is not a right angle, the area is less, decreasing to zero as $\mathbf{A}$ and $\mathbf{B}$ become (anti-)parallel.
A: @Mehrdad
I had the same question as you have expressed:

I feel like the hard part is understanding why dot product has
anything to do with projection in the first place. Why does the sum of
a componentwise product tell you something about the vectors'
projection?

After some thinking, I come up with the following reasoning, not sure if it make sense to you.


*

*In Section 6.5-1 of Lathi’s Linear Systems and Signals, 2nd, projection of $\mathbf{x}$ along $\mathbf{y}$ can be interpreted as a way to minimize the "error" $\mathbf{e}$, when $\mathbf{x}$ is expressed as $\mathbf{x}=c\mathbf{y}+\mathbf{e}$, where $c\mathbf{y}$ is the component of $\mathbf{a}$ in the direction of $\mathbf{b}$, and $\mathbf{e}$ is the "error" vector, which has the minimum length when it's perpendicular to $\mathbf{b}$.


*Btw: the "error" vector gives hints on how much $\mathbf{x}$ is differ from $c\mathbf{y}$. In this sense, projection is a way to find similarities of two vectors, and this may explain why correlation is calculated exactly in the same way as dot product.


*Now the question can be rephrased as: why product has anything to do with correlation/similarity? Let's take plain numbers (not vectors) for now. To find difference between two numbers $a$ and $b$, we do subtraction $a-b$, which can be either positive or negative. Most of the time people only cares about the absolute value of the difference $|a-b|$, but $|a-b|$ is not convenient in mathematical manipulation. So people square the difference, $(a-b)^2$...this is where the product comes in. The formula $(a-b)^2=a^2+b^2-2ab$ suggests that the product $ab$ has its place as a measure of the difference or similarity between $a$ and $b$.
