柯西-许瓦尔兹不等式

Consider the two n-dimensional vectors: $$\vec{a}= \left< a_1,a_2,...,a_n \right>\text{ and } \vec{b}=\left< b_1,b_2,...,b_n \right>.$$

A simple calculation shows $$|\vec{a}|^2|\vec{b}|^2=\left( a_1^2+a_2^2+...+a_n^2\right)\left( b_1^2+b_2^2+...+b_n^2\right).$$

In addition: $\left(\vec{a}\cdot\vec{b}\right)^2=\left(a_1b_1+a_2b_2+...+a_nb_n\right)^2$.

Let $\theta$ be the angle between vectors $\vec{a}$ and $\vec{b}$ when the two are placed with their initial points at the same location.

We know that $\left( \vec{a}\cdot\vec{b} \right) = \lvert \vec{a}\rvert \lvert \vec{b} \rvert \cos\theta$.

Since $-1\le \cos\theta \le1$, $0\le \cos^2\theta \le1$.

Therefore, $$\left( \vec{a}\cdot\vec{b} \right) ^2= \lvert \vec{a}\rvert ^2 \lvert \vec{b} \rvert^2 \cos^2\theta \le \lvert \vec{a}\rvert^2 \lvert \vec{b}\rvert^2.$$

Plugging in our earlier expressions yields the Cauchy-Schwarz Inequality: $$\left( a_1^2+a_2^2+...+a_n^2 \right)\left(b_1^2+b_2^2+...+b_n^2 \right) \ge \left(a_1b_1+a_2b_2+...+a_nb_n\right)^2.$$

The vectors also give us a very nice way of seeing when equality holds.

Clearly, equality holds if and only if $\cos^2 \theta=1$, which means that the vectors $\vec{a}$ and $\vec{b}$ are parallel.

This happens if and only if there is a real number $k$ such that $k\vec{a}=\vec{b}$, or if one or both of the vectors are the zero vector.

Thus, equality holds if and only if one or more of the following conditions hold (we must make a special case to avoid division by zero): $$b_1=b_2=...=b_n=0$$ $$\frac{a_1}{b_1}=\frac{a_2}{b_2}=...=\frac{a_n}{b_n}$$

Here is another (and very simple) proof for the Cauchy-Schwartz Inequality.

Consider:

$(a_1x-b_1)^2=a_1^2x^2-2a_1b_1x+b_1^2\ge 0$

$(a_2x-b_2)^2=a_2^2x^2-2a_2b_2x+b_2^2\ge0$

$(a_3x-b_3)^2=a_3^2x^2-2a_3b_3x+b_3^2\ge0$

$...$

$(a_nx-b_n)^2=a_n^2x^2-2a_nb_nx+b_n^2\ge0$

Add all of the above inequalities and we get:

$(a_1^2+a_2^2+...+a_n^2)x^2-2(a_1b_1+a_2b_2+...+a_nb_n)x+(b_1^2+b_2^2+...+b_n^2)?0$

Above, we have a quadratic inequality in the form $ax^2+bx+c\ge0$

On a graph, this will look something like these:

This is because $(a_1^2+a_2^2+...+a_n^2)x^2-2(a_1b_1+a_2b_2+...+a_nb_n)x+(b_1^2+b_2^2+...+b_n^2)$ will always be greater or equal to 0. In these cases, since the graph either intersects the x-axis at one point (has one solution) or does not intersect it at all (no solution), we can use the discriminant $b^2-4ac\le0$ where

$b=-2(a_1b_1+a_2b_2+...+a_nb_n), a=(a_1^2+a_2^2+...+a_n^2), c=(b_1^2+b_2^2+...+b_n^2)$

Thus, $4(a_1b_1+a_2b_2+...+a_nb_n)^2-4(a_1^2+a_2^2+...+a_n^2)(b_1^2+b_2^2+...+b_n^2)\le0$

$(a_1b_1+a_2b_2+...+a_nb_n)^2\le(a_1^2+a_2^2+...+a_n^2)(b_1^2+b_2^2+...+b_n^2)$

Finally, we have the original Cauchy-Schwartz inequality!