11. Cauchy–Schwarz Inequality

Where:

• $\overrightarrow{𝑢}\cdot \overrightarrow{𝑣}$: dot product of vectors $\overrightarrow{𝑢}$ and $\overrightarrow{𝑣}$
• $\Vert \overrightarrow{𝑢}\Vert$ and $\Vert \overrightarrow{𝑣}\Vert$: magnitudes (lengths) of vectors $\overrightarrow{𝑢}$ and $\overrightarrow{𝑣}$

Step 1: Understand the dot product

The dot product of two vectors $\overrightarrow{𝑢}=\left[{𝑢}_1,{𝑢}_2,\dots ,{𝑢}_{𝑛}\right]$ and $\overrightarrow{𝑣}=\left[{𝑣}_1,{𝑣}_2,\dots ,{𝑣}_{𝑛}\right]$ is calculated as:

The magnitude (or norm) of a vector $\overrightarrow{𝑢}$ is:

Step 2: Define a new function

We introduce a parameter $𝑡\in ℝ$ and define a new vector:

Now, consider the dot product of this new vector with itself, which is always non-negative because it represents the square of the magnitude of $𝑤\left(𝑡\right)$:

This inequality makes sense because the dot product of any vector with itself is the square of its magnitude, and a square is always non-negative.

Step 3: Expand the dot product

Expand $𝑤\left(𝑡\right)\cdot 𝑤\left(𝑡\right)$:

Now, we apply the distributive property of the dot product, which behaves similarly to the distributive property of multiplication. We expand each term:

Since the dot product is commutative ($\overrightarrow{𝑢}\cdot \overrightarrow{𝑣}=\overrightarrow{𝑣}\cdot \overrightarrow{𝑢}$), we can rewrite this as:

This simplifies to:

We’ve now expressed the result of expanding the dot product as a quadratic expression in $𝑡$, where

• $\Vert \overrightarrow{𝑢}{\Vert }^2$ is a constant term,
• $-2𝑡\left(\overrightarrow{𝑢}\cdot \overrightarrow{𝑣}\right)$ is the linear term in $𝑡$
• ${𝑡}^2\Vert \overrightarrow{𝑣}{\Vert }^2$ is the quadratic term

Step 4: Treat as a quadratic equation

Now that we have the quadratic expression:

We recognize this as a standard quadratic inequality of the form $𝑎{𝑡}^2+𝑏𝑡+𝑐\ge 0$, where:

• $𝑎=\Vert \overrightarrow{𝑣}{\Vert }^2$
• $𝑏=-2\left(\overrightarrow{𝑢}\cdot \overrightarrow{𝑣}\right)$
• $𝑐=\Vert \overrightarrow{𝑢}{\Vert }^2$

For any quadratic expression $𝑎{𝑡}^2+𝑏𝑡+𝑐$ to always be non-negative, its discriminant must be less than or equal to zero. The discriminant of a quadratic equation $𝑎{𝑡}^2+𝑏𝑡+𝑐=0$ is given by:

Substituting in the values of $𝑎$, $𝑏$, and $𝑐$ from our expression:

Simplifying:

Step 5: Apply the discriminant condition

For the quadratic inequality to hold, the discriminant must be less than or equal to zero:

Divide by 4:

Take the square root of both sides: