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Proof of the Cauchy-Schwarz Inequality: $|\mathbb{x} \cdot \mathbb{y}| \leq |\mathbb{x}| |\mathbb{y}|$

Proof

We will prove this by induction.

Base case $\mathbb{R}^2$

Let $\mathbb{x}$, $\mathbb{y} \in \mathbb{R}^2$ be $(x_1, x_2)$ and $(y_1, y_2)$ respectively.

Start with $|\mathbb{x} \cdot \mathbb{y}|$ and note that:

$$\begin{align} |\mathbb{x} \cdot \mathbb{y}|^2 &= x_1^2 y_1^2 + 2 x_1 x_2 y_1 y_2 + x_2^2 y_2^2 \\ &\leq (x_1^2 y_1^2 + 2 x_1 x_2 y_1 y_2 + x_2^2 y_2^2) + (x_1y_2 - x_2y_1)^2 \\ &= x_1^2 y_1^2 + x_1^2 y_2^2 + x_1^2 y_2^2 + x_2^2 y_2^2 \\ &= (x_1^2 + x_2^2)(y_1^2+ y_2^2) \\ &= (|x||y|)^2 \end{align}$$

Since both $|\mathbb{x} \cdot \mathbb{y}|$ and $|x||y|$ are positive and the square root is an increasing function, we have:

$$|\mathbb{x} \cdot \mathbb{y}| \leq |x||y|$$

which proves the base case.

Induction Step:

Let $\mathbb{x}$, $\mathbb{y} \in \mathbb{R}^n$ be $(x_1, \dots, x_n)$ and $(y_1, \dots, y_n)$ respectively.

Assume $|\mathbb{x} \cdot \mathbb{y}| \leq |\mathbb{x}||\mathbb{y}|$. We will then show that this also works for $\mathbb{x'}$, $\mathbb{y'} \in \mathbb{R}^{n+1}$, where $(x_1, \dots, x_n, x_{n+1})$ and $(y_1, \dots, y_n, y_{n+1})$ respectively

By our assumption:

$$\begin{align} |\mathbb{x} \cdot \mathbb{y}| &\leq |x||y| \\ \sqrt{(\mathbb{x} \cdot \mathbb{y})^2} &\leq \sqrt{\mathbb{x} \cdot \mathbb{x}} \sqrt{\mathbb{y} \cdot \mathbb{y}} \\ \sqrt{(x_1y_1 + \dots x_ny_n)^2} &\leq \sqrt{x_1^2 + \dots + x_n^2} \sqrt{y_1^2 + \dots + y_n^2} \\ \end{align}$$

Consider $|\mathbb{x'} \cdot \mathbb{y'}|^2$:

$$\begin{align*} &= \sqrt{(\mathbb{x'} \cdot \mathbb{y'})^2}^2\\ &= (\mathbb{x'} \cdot \mathbb{y'})^2\\ &= (x_1y_1 + \dots x_ny_n +x_{n+1}y_{n+1})^2 \\ &= (\mathbb{x} \cdot \mathbb{y} +x_{n+1}y_{n+1})^2 \\ &= (\mathbb{x} \cdot \mathbb{y})^2 + 2x_{n+1}y_{n+1}(\mathbb{x} \cdot \mathbb{y}) + (x_{n+1}y_{n+1})^2 \end{align*}$$

And $(|\mathbb{x'}||\mathbb{y'}|)^2$:

$$\begin{align*} &= (\sqrt{\mathbb{x'} \cdot \mathbb{x'}} \sqrt{\mathbb{y'} \cdot \mathbb{y'}})^2 \\ &= (\mathbb{x'} \cdot \mathbb{x'})(\mathbb{y'} \cdot \mathbb{y'}) \\ &=(x_1^2 + \dots + x_n^2 + x_{n+1}^2)(y_1^2 + \dots + y_n^2 + y_{n+1}^2) \\ &= (\mathbb{x} \cdot \mathbb{x} + x_{n+1}^2)(\mathbb{y} \cdot \mathbb{y} + y_{n+1}^2) \\ &= (\mathbb{x} \cdot \mathbb{x})(\mathbb{y} \cdot \mathbb{y}) + y_{n+1}^2(\mathbb{x} \cdot \mathbb{x}) + x_{n+1}^2(\mathbb{y} \cdot \mathbb{y}) + x_{n+1}^2y_{n+1}^2 \end{align*}$$

Note that:

$$\begin{align*} |\mathbb{x'} \cdot \mathbb{y'}|^2 &\leq |\mathbb{x'} \cdot \mathbb{y'}|^2 + |(y_{n+1}\mathbb{x} - x_{n+1}\mathbb{y})|^2 \\ & = |\mathbb{x'} \cdot \mathbb{y'}|^2 + y_{n+1}^2(\mathbb{x} \cdot \mathbb{x}) -2x_{n+1}y_{n+1}(\mathbb{x} \cdot \mathbb{y}) + x_{n+1}^2(\mathbb{y} \cdot \mathbb{y}) \\ & = (\mathbb{x} \cdot \mathbb{y})^2 + 2x_{n+1}y_{n+1} (\mathbb{x} \cdot \mathbb{y})+ (x_{n+1}y_{n+1})^2 + y_{n+1}^2(\mathbb{x} \cdot \mathbb{x}) -2x_{n+1}y_{n+1}(\mathbb{x} \cdot \mathbb{y}) + x_{n+1}^2(\mathbb{y} \cdot \mathbb{y}) \\ &= (\mathbb{x} \cdot \mathbb{y})^2 + (x_{n+1}y_{n+1})^2 + y_{n+1}^2(\mathbb{x} \cdot \mathbb{x}) + x_{n+1}^2(\mathbb{y} \cdot \mathbb{y}) \\ &= (\mathbb{x} \cdot \mathbb{x})(\mathbb{y} \cdot \mathbb{y}) + y_{n+1}^2(\mathbb{x} \cdot \mathbb{x}) + x_{n+1}^2(\mathbb{y} \cdot \mathbb{y}) + x_{n+1}^2y_{n+1}^2 \\ &= (|\mathbb{x'}||\mathbb{y'}|)^2 \end{align*}$$

But since the sqaure root is an increasing function and both $|\mathbb{x'} \cdot \mathbb{y'}|$ and $|\mathbb{x'}||\mathbb{y'}|$ are positive, $|\mathbb{x'} \cdot \mathbb{y'}|^2 \leq (|\mathbb{x'}||\mathbb{y'}|)^2$ means that $|\mathbb{x'} \cdot \mathbb{y'}| \leq |\mathbb{x'}||\mathbb{y'}|$

So we have shown that if we assume $|\mathbb{x} \cdot \mathbb{y}| \leq |\mathbb{x}||\mathbb{y}|$ for $\mathbb{x}, \mathbb{y} \in \mathbb{R}^n$ then $|\mathbb{x'} \cdot \mathbb{y'}| \leq |\mathbb{x'}||\mathbb{y'}|$ for $\mathbb{x'}, \mathbb{y'} \in \mathbb{R}^{n+1}$ as desired.

By induction, our proof is finished.