In $\mathbb{R}^n$: $|x \cdot y| \leq \|x\| \cdot \|y\|$, or equivalently:

$\left(\sum_{i=1}^n x_i y_i\right)^2 \leq \left(\sum_{i=1}^n x_i^2\right) \left(\sum_{i=1}^n y_i^2\right).$

For $x = (1, 2, 3)$, $y = (4, 5, 6)$: $|x \cdot y| = |4 + 10 + 18| = 32$ and $\|x\| \cdot \|y\| = \sqrt{14} \cdot \sqrt{77} = \sqrt{1078} \approx 32.83$. So $32 \leq 32.83$ ✓.

For continuous functions on $[a, b]$:

$\left|\int_a^b f(x)g(x)\,dx\right|^2 \leq \int_a^b f(x)^2\,dx \cdot \int_a^b g(x)^2\,dx.$

For $f(x) = x$ and $g(x) = 1$ on $[0, 1]$: $|\int_0^1 x\,dx|^2 = \frac{1}{4}$ and $\int_0^1 x^2\,dx \cdot \int_0^1 1\,dx = \frac{1}{3}$. So $\frac{1}{4} \leq \frac{1}{3}$ ✓.

For real numbers $a_1, \ldots, a_n$ and $b_1, \ldots, b_n$:

$(a_1 b_1 + \cdots + a_n b_n)^2 \leq (a_1^2 + \cdots + a_n^2)(b_1^2 + \cdots + b_n^2).$

For $n = 2$: $(a_1 b_1 + a_2 b_2)^2 \leq (a_1^2 + a_2^2)(b_1^2 + b_2^2)$. Expanding: $a_1^2 b_1^2 + 2a_1 b_1 a_2 b_2 + a_2^2 b_2^2 \leq a_1^2 b_1^2 + a_1^2 b_2^2 + a_2^2 b_1^2 + a_2^2 b_2^2$. The difference is $(a_1 b_2 - a_2 b_1)^2 \geq 0$ ✓.

Equality in $|\langle u, v \rangle| = \|u\| \|v\|$ iff $u = \alpha v$ or $v = 0$.

$u = (2, 4)$, $v = (1, 2)$: $|\langle u, v \rangle| = |2 + 8| = 10$ and $\|u\| \|v\| = \sqrt{20} \cdot \sqrt{5} = 10$. Equality holds since $u = 2v$.

$u = (1, 0)$, $v = (0, 1)$: $|\langle u, v \rangle| = 0$ and $\|u\| \|v\| = 1$. Strict inequality because $u$ and $v$ are linearly independent.

The Cauchy--Schwarz inequality is equivalent to $|\cos\theta| \leq 1$ where $\theta$ is the angle between $u$ and $v$:

$\cos\theta = \frac{\langle u, v \rangle}{\|u\| \|v\|} \in [-1, 1].$

Equality ($|\cos\theta| = 1$) means $\theta = 0$ or $\theta = \pi$, i.e., the vectors are parallel.

$u = (3, 0)$, $v = (0, 4)$: $\|u + v\| = 5 \leq 3 + 4 = 7$.

$u = (1, 2)$, $v = (3, 6)$: $\|u + v\| = \|(4, 8)\| = 4\sqrt{5}$ and $\|u\| + \|v\| = \sqrt{5} + 3\sqrt{5} = 4\sqrt{5}$. Equality because $v = 3u$.

Also from Cauchy--Schwarz: $|\|u\| - \|v\|| \leq \|u - v\|$.

$u = (5, 0)$, $v = (3, 4)$: $|\|u\| - \|v\|| = |5 - 5| = 0 \leq \|(2, -4)\| = 2\sqrt{5}$ ✓.

In probability, the Cauchy--Schwarz inequality applied to $\langle X, Y \rangle = E[XY]$ (with mean-centered random variables) gives:

$|E[XY]|^2 \leq E[X^2] \cdot E[Y^2],$

or equivalently $|\operatorname{Cov}(X, Y)| \leq \sigma_X \sigma_Y$. The correlation coefficient $\rho = \frac{\operatorname{Cov}(X,Y)}{\sigma_X \sigma_Y}$ satisfies $|\rho| \leq 1$ by Cauchy--Schwarz.

Apply Cauchy--Schwarz to $a = (a_1, \ldots, a_n)$ and $b = (1, 1, \ldots, 1)$:

$(a_1 + \cdots + a_n)^2 \leq n(a_1^2 + \cdots + a_n^2),$

i.e., $\left(\frac{a_1 + \cdots + a_n}{n}\right)^2 \leq \frac{a_1^2 + \cdots + a_n^2}{n}$, which says AM $\leq$ QM (arithmetic mean at most quadratic mean).

For $a = (1, 2, 3)$: $\text{AM} = 2$, $\text{QM} = \sqrt{14/3} \approx 2.16$. So $2 \leq 2.16$ ✓.

For sequences $(a_n)$ and $(b_n)$ with $\sum a_n^2 < \infty$ and $\sum b_n^2 < \infty$:

$\left(\sum_{n=1}^\infty a_n b_n\right)^2 \leq \sum_{n=1}^\infty a_n^2 \cdot \sum_{n=1}^\infty b_n^2.$

For $a_n = 1/n$ and $b_n = 1/n^2$: $\sum a_n b_n = \sum 1/n^3 = \zeta(3) \approx 1.202$, $\sum a_n^2 = \pi^2/6$, $\sum b_n^2 = \pi^4/90$. So $\zeta(3)^2 \approx 1.44 \leq \frac{\pi^2}{6} \cdot \frac{\pi^4}{90} \approx 1.78$ ✓.

The maximum of $\langle u, v \rangle$ subject to $\|u\| = 1$ is $\|v\|$, achieved when $u = v / \|v\|$.

This follows directly from Cauchy--Schwarz: $\langle u, v \rangle \leq |\langle u, v \rangle| \leq \|u\| \|v\| = \|v\|$.

Example: maximize $x + 2y + 3z$ subject to $x^2 + y^2 + z^2 = 1$. This is $\langle (x,y,z), (1,2,3) \rangle$ with $\|(x,y,z)\| = 1$. Maximum is $\|(1,2,3)\| = \sqrt{14}$, achieved at $(x,y,z) = \frac{1}{\sqrt{14}}(1,2,3)$.

For a matrix $A$ with columns $a_1, \ldots, a_n$, repeated application of Cauchy--Schwarz (via Gram--Schmidt) gives Hadamard's inequality:

$|\det A| \leq \|a_1\| \cdot \|a_2\| \cdots \|a_n\|.$

Equality holds iff the columns are orthogonal. This is because Gram--Schmidt gives $A = QR$ with $|\det A| = |\det R| = r_{11} \cdots r_{nn}$ and $r_{jj} = \|u_j\| \leq \|v_j\| = \|a_j\|$.