柯西不等式的证明

$$\sum_{i=1}^{n} a_{i}^{2} \sum_{i=1}^{n} b_{i}^{2} \ge (\sum_{i=1}^{n} a_{i} b_{i})^{2} $$$$(a_{1}^{2} + a_{2}^{2} + \cdots + a_{n}^{2})(b_{1}^{2} + b_{2}^{2} + \cdots + b_{n}^{2}) \ge (a_{1}b_{1} + a_{2}b_{2} + \cdots + a_{n}b_{n})^2$$

利用 $b^{2} - 4ac \ge 0$

证明：构造系列函数：

$$f_{1}(x) = a_{1}^{2}x^{2} + 2a_{1}b_{1}x + b_{1}^{2} = (a_{1}x+b_{1})^{2} \ge 0$$$$f_{2}(x) = a_{2}^{2}x^{2} + 2a_{2}b_{2}x + b_{2}^{2} = (a_{2}x+b_{2})^{2} \ge 0$$$$\cdots \cdots$$$$f_{n}(x) = a_{n}^{2}x^{2} + 2a_{n}b_{n}x + b_{n}^{2} = (a_{n}x+b_{n})^{2} \ge 0$$$$g(x) = f_{1}(x) + f_{2}(x) + \cdots + f_{n}(x)$$$$\therefore g(x) = (a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2})x^{2}+2(a_{1}b_{1}+a_{2}b_{2}+\cdots+a_{n}b_{n}) \ge 0$$$$\Delta_{x} \le 0$$$$\therefore 4(a_{1}b_{1} + a_{2}b_{2} + \cdots + a_{n}b_{n})^2 - 4(a_{1}^{2} + a_{2}^{2} + \cdots + a_{n}^{2})(b_{1}^{2} + b_{2}^{2} + \cdots + b_{n}^{2}) \le 0$$$$\therefore (a_{1}^{2} + a_{2}^{2} + \cdots + a_{n}^{2})(b_{1}^{2} + b_{2}^{2} + \cdots + b_{n}^{2}) \ge (a_{1}b_{1} + a_{2}b_{2} + \cdots + a_{n}b_{n})^2$$

当上式取等时，$\Delta = 0$，有 $x$ 使得 $g(x)=0$ 成立
则 $x = - \frac {b_{1}} {a_{1}} = - \frac {b_{2}} {a_{2}} = \cdots = - \frac {b_{n}} {a_{n}}$
即当且仅当 $\frac {b_{1}} {a_{1}} = \frac {b_{2}} {a_{2}} = \cdots = \frac {b_{n}} {a_{n}}$ 时等号成立

$$(a_{1}^{2} + a_{2}^{2} + \cdots + a_{n}^{2})(b_{1}^{2} + b_{2}^{2} + \cdots + b_{n}^{2}) \ge (a_{1}b_{1} + a_{2}b_{2} + \cdots + a_{n}b_{n})^2$$

当且仅当 $\frac {b_{1}} {a_{1}} = \frac {b_{2}} {a_{2}} = \cdots = \frac {b_{n}} {a_{n}}$ 时等号成立