In my opinion, the following proof is the shortest, simplest and best proof of the Cauchy-Schwarz inequality. It is a proof developed in The Cauchy-Schwarz Master Class: An Introduction to the Art of Mathematical Inequalities [1]. Below are three variations of the proof at three increasing levels of abstraction. These three variations are expressed respectively in terms of:

• random variables [2]

• vectors of any real inner product space [1]

• vectors of any inner product space (real or complex) [3]

The Cauchy-Schwarz inequality was originally expressed in terms of sequences of numbers [1]. The continuous analogue is in terms of two integrable functions [4].

In terms of random variables

Given any two random variables $X$ and $Y$, $${\operatorname{E}\!\left[{ XY}\right]}^2 \le {\operatorname{E}\!\left[{ X^2}\right]} {\operatorname{E}\!\left[{ Y^2}\right]}$$ with equality holding iff $a X + b Y = 0$ for some constants $a$ and $b$, at least one non-zero (i.e. $X$ and $Y$ are linearly dependent).

Proof

If either ${\operatorname{E}\!\left[{ X^2}\right]} = 0$ or ${\operatorname{E}\!\left[{ Y^2}\right]} = 0$ then ${\operatorname{E}\!\left[{ XY}\right]} = 0$. Otherwise define $$\hat{X} := \frac{X}{ \sqrt{{\operatorname{E}\!\left[{ X^2}\right]}} } \; \text{ and } \; \hat{Y} := \frac{Y}{ \sqrt{{\operatorname{E}\!\left[{ Y^2}\right]}} }$$ for which ${\operatorname{E}\!\left[{ \hat{X}^2}\right]} = {\operatorname{E}\!\left[{ \hat{Y}^2}\right]} = 1$. The proof follows from the product of two numbers always being less than or equal to the average of their squares $$\begin{aligned} 0 & \le {\operatorname{E}\!\left[{ (\hat{X} - \hat{Y})^2 }\right]} \\ {\operatorname{E}\!\left[{ \hat{X} \hat{Y} }\right]} & \le {\operatorname{E}\!\left[{ \frac{ \hat{X}^2 + \hat{Y}^2 }{2} }\right]} \\ \frac{{\operatorname{E}\!\left[{ XY}\right]}}{ \sqrt{{\operatorname{E}\!\left[{ X^2}\right]}} \sqrt{{\operatorname{E}\!\left[{ Y^2}\right]}} } & \le \frac{1+1}{2} \\ {\operatorname{E}\!\left[{ XY}\right]}^2 & \le {\operatorname{E}\!\left[{ X^2}\right]} {\operatorname{E}\!\left[{ Y^2}\right]} \end{aligned}$$ If both sides of the inequality are equal, linear dependence follows since either $X = 0$ or $Y = 0$ or $$\frac{1}{\sqrt{{\operatorname{E}\!\left[{ X^2}\right]}}} X + \frac{1}{\sqrt{{\operatorname{E}\!\left[{ Y^2}\right]}}} Y = \hat{X} - \hat{Y} = 0$$ If $X$ and $Y$ are linearly dependent, either $X = k Y$ or $Y = k X$ for some constant $k$, either way both sides of the inequality are equal.

QED

In terms of vectors of a real inner product space

The probabilistic proof can be generalized to any real inner product space as shown in [1].

Given any vectors $x, y$ from a real inner product space, the Cauchy-Schwarz inequality is $$\left\langle{ x}, { y}\right\rangle \le \left\|{ x}\right\| \left\|{ y}\right\|$$ with equality holding iff $x$ and $y$ are linearly dependent.

Proof

If either $\left\|{ x}\right\| = 0$ or $\left\|{ y}\right\| = 0$ then $\left\langle{ x}, { y}\right\rangle = 0$. Otherwise define $$\hat{x} := \frac{x}{ \left\|{ x}\right\| } \; \text{ and } \; \hat{y} := \frac{y}{ \left\|{ y}\right\| }$$ for which $\left\|{ \hat{x}}\right\| = \left\|{ \hat{y}}\right\| = 1$. $$\begin{aligned} 0 & \le \left\langle{ \hat{x} - \hat{y}}, { \hat{x} - \hat{y}}\right\rangle \\ 2 \left\langle{ \hat{x}}, { \hat{y}}\right\rangle & \le \left\langle{ \hat{x}}, { \hat{x}}\right\rangle + \left\langle{ \hat{y}}, { \hat{y}}\right\rangle \\ 2 \; \frac{ \left\langle{ x}, { y}\right\rangle }{ \left\|{ x}\right\| \left\|{ y}\right\| } & \le 1+1 \\ \left\langle{ x}, { y}\right\rangle & \le \left\|{ x}\right\| \left\|{ y}\right\| \end{aligned}$$ If both sides of the inequality are equal, linear dependence follows since either $x = \vec{0}$ or $y = \vec{0}$ or $$\frac{1}{\left\|{ x}\right\|} x + \frac{1}{\left\|{ y}\right\|} y = \hat{x} - \hat{y} = \vec{0}$$ If $x$ and $y$ are linearly dependent, either $x = \lambda y$ or $y = \lambda x$ for some scaler $\lambda$, either way both sides of the inequality are equal.

QED

In terms of vectors of an inner product space

This section considers the Cauchy-Schwarz inequality for vectors of a real or complex inner product space.

The proof for real inner produce spaces does not work for complex inner product spaces because $\left\langle{ y}, { x}\right\rangle = \overline{\left\langle{ x}, { y}\right\rangle}$ (complex conjugate).

The proof is effectively the same as the previous proof for real inner product spaces. But the normalized vectors $\hat{x}$ and $\hat{y}$ must be “rotated” in the complex plane so that both sides of the inequality remain real. This rotation will be done via a multiplier $\alpha$.

Proof

Let $\hat{x}$ and $\hat{y}$ be defined as in the proof for real inner product spaces. If $\left\langle{ \hat{x}}, { \hat{y}}\right\rangle = 0$ the inequality holds, otherwise let $$\alpha := \sqrt{ \frac{\left\langle{ \hat{y}}, { \hat{x}}\right\rangle}{\left|{ \left\langle{ \hat{x}}, { \hat{y}}\right\rangle}\right|} }$$ for which the following convenient properties hold $$\alpha \overline{\alpha} = 1 = \overline{\alpha} \alpha$$ $$\alpha^2 \left\langle{ \hat{x}}, { \hat{y}}\right\rangle = \frac{\left\langle{ \hat{y}}, { \hat{x}}\right\rangle\left\langle{ \hat{x}}, { \hat{y}}\right\rangle}{\left|{ \left\langle{ \hat{x}}, { \hat{y}}\right\rangle}\right|} = \frac{\left|{ \left\langle{ \hat{x}}, { \hat{y}}\right\rangle}\right|^2}{\left|{ \left\langle{ \hat{x}}, { \hat{y}}\right\rangle}\right|} = \left|{ \left\langle{ \hat{x}}, { \hat{y}}\right\rangle}\right| = \overline{\alpha}^2 \left\langle{ \hat{y}}, { \hat{x}}\right\rangle$$

The proof proceeds like with a real inner product space but using $\alpha$, $$\begin{aligned} 0 & \le \left\langle{ \alpha \hat{x} - \overline{\alpha} \hat{y}}, { \alpha \hat{x} - \overline{\alpha} \hat{y}}\right\rangle \\ & = \alpha \overline{\alpha} \left\langle{ \hat{x}}, { \hat{x}}\right\rangle - \alpha^2 \left\langle{ \hat{x}}, { \hat{y}}\right\rangle - \overline{\alpha}^2 \left\langle{ \hat{y}}, { \hat{x}}\right\rangle + \overline{\alpha} \alpha \left\langle{ \hat{y}}, { \hat{y}}\right\rangle \\ 2 \left|{ \left\langle{ \hat{x}}, { \hat{y}}\right\rangle}\right| & \le \left\langle{ \hat{x}}, { \hat{x}}\right\rangle + \left\langle{ \hat{y}}, { \hat{y}}\right\rangle \\ 2 \; \frac{ \left|{ \left\langle{ x}, { y}\right\rangle}\right| }{ \left\|{ x}\right\| \left\|{ y}\right\| } & \le 1+1 \\ \left|{ \left\langle{ x}, { y}\right\rangle}\right| & \le \left\|{ x}\right\| \left\|{ y}\right\| \end{aligned}$$ If both sides of the inequality are equal, linear dependence follows since either $x = \vec{0}$ or $y = \vec{0}$ or $$\frac{\alpha}{\left\|{ x}\right\|} x + \frac{\overline{\alpha}}{\left\|{ y}\right\|} y = \alpha \hat{x} - \overline{\alpha} \hat{y} = \vec{0}$$ If $x$ and $y$ are linearly dependent, either $x = \lambda y$ or $y = \lambda x$ for some scaler $\lambda$, either way both sides of the inequality are equal.

QED

References