$\bullet$ 二維形式的柯西不等式：
$$(a^{2} + b^{2})(c^{2} + d^{2}) \geq (ac + bd)^{2}$$
當且僅當 $ad = bc$ 時等號成立。
$\bullet$ 三維形式的柯西不等式：
$$(a_{1}^{2} + a_{2}^{2} + a_{3}^{2})(b_{1}^{2} + b_{2}^{2} + b_{3}^{2}) \geq (a_{1}b_{1} + a_{2}b_{2} + a_{3}b_{3})^{2}$$
當且僅當 $\frac{a_{1}}{b_{1}} = \frac{a_{2}}{b_{2}} = \frac{a_{3}}{b_{3}}$ 時等號成立。
$\bullet$ 一般形式的柯西不等式：
$$(a_{1}^{2} + a_{2}^{2} + \cdots + a_{n}^{2})(b_{1}^{2} + b_{2}^{2} + \cdots + b_{n}^{2}) \geq (a_{1}b_{1} + a_{2}b_{2} + \cdots + a_{n}b_{n})^{2} \\
\Rightarrow \sum_{i=1}^{n}a_{i}^{2}\sum_{i=1}^{n}b_{i}^{2} \geq \left ( \sum_{i=1}^{n}a_{i}b_{i} \right )^{2}$$
當且僅當 $\frac{a_{1}}{b_{1}} = \frac{a_{2}}{b_{2}} = \cdots = \frac{a_{n}}{b_{n}}$ 時等號成立。
式子的左邊是兩個平方和的乘積，式子右邊是交叉項和的平方。
證明：
定義函數 $f(x)$ 為
$$f(x) = (a_{1} + b_{1}x)^{2} + (a_{2} + b_{2}x)^{2} + \cdots + (a_{n} + b_{n}x)^{2}$$
將 $f(x)$ 轉化為二元函數的標准形式 $y = ax^{2} + bx + c$ 得
$$f(x) = (b_{1}^{2} + b_{2}^{2} + \cdots + b_{n}^{2})x^{2} + 2(a_{1}b_{1} + a_{2}b_{2} + \cdots + a_{n}b_{n})x + (a_{1}^{2} + a_{2}^{2} + \cdots + a_{n}^{2})$$
因為 $f(x) \geq 0$，所以它只有一個解或無解，即
$$\Delta = 4(a_{1}b_{1} + a_{2}b_{2} + \cdots + a_{n}b_{n})^{2} - 4(b_{1}^{2} + b_{2}^{2} + \cdots + b_{n}^{2})(a_{1}^{2} + a_{2}^{2} + \cdots + a_{n}^{2}) \leq 0$$
所以
$$(a_{1}^{2} + a_{2}^{2} + \cdots + a_{n}^{2})(b_{1}^{2} + b_{2}^{2} + \cdots + b_{n}^{2}) \geq (a_{1}b_{1} + a_{2}b_{2} + \cdots + a_{n}b_{n})^{2}$$
令函數 $f(x) = 0$，則每個平方項都必須為 $0$，即
$$a_{1} + b_{1}x = 0 \Rightarrow x = -\frac{a_{1}}{b_{1}} \\
a_{2} + b_{2}x = 0 \Rightarrow x = -\frac{a_{2}}{b_{2}} \\
\vdots \\
a_{n} + b_{n}x = 0 \Rightarrow x = -\frac{a_{n}}{b_{n}}$$
則要使函數有零點，即 $\Delta = 0$，則必須有
$$\frac{a_{1}}{b_{1}} = \frac{a_{2}}{b_{2}} = \cdots = \frac{a_{n}}{b_{n}}$$
證畢