Table of Contents
对于等式 $ \left(a_ {1}^{2} + a_ {2}^{2}\right) \left(b_ {1}^{2} + b_ {2}^{2}\right) = \left(a_ {1}b_ {1} + a_ {2}b_ {2}\right)^{2} + \left(a_ {1}b_ {2} - a_ {2}b_ {1}\right)^{2} $，当
$ \begin{equation} a_ {1} = \lambda b_ {1} 且 a_ {2} = \lambda b_ {2} \qquad 对某些实数 \lambda \end{equation} $
时，$ \left(a_ {1}b_ {2} - a_ {2}b_ {1}\right)^{2} = 0 $。
PASSAGE TO A MORE GENERAL IDENTITY
对于更一般化的公式，我们设 $ Q_ {n} $为
$ \begin{equation} \left(a_ {1}^{2} + a_ {2}^{2} + \cdots + a_ {n}^{2}\right) \left(b_ {1}^{2} + b_ {2}^{2} + \cdots + b_ {n}^{2}\right) - \left(a_ {1}b_ {1} + a_ {2}b_ {2} + \cdots + a_ {n}b_ {n}\right)^{2} \end{equation} $
我们可简化重写为
$ \begin{equation} Q_ {n} = \sum_ {i=1}^{n} \sum_ {j=1}^{n}a_ {i}^{2}b_ {j}^{2} - \sum_ {i=1}^{n} \sum_ {j=1}^{n}a_ {i}b_ {i}a_ {j}b_ {j} \end{equation} $
为使得等式更有对称性，我们再重写为
$ \begin{equation} Q_ {n} = \frac{1}{2}\sum_ {i=1}^{n} \sum_ {j=1}^{n}\left(a_ {i}^{2}b_ {j}^{2} + a_ {j}^{2}b_ {i}^{2}\right) - \sum_ {i=1}^{n} \sum_ {j=1}^{n}a_ {i}b_ {i}a_ {j}b_ {j} \end{equation} $
我们分解一下，有
$ \begin{equation} Q_ {n} = \frac{1}{2}\sum_ {i=1}^{n} \sum_ {j=1}^{n} \left\{ a_ {i}^{2}b_ {j}^{2} - 2a_ {i}b_ {j}a_ {j}b_ {i} + a_ {j}^{2}b_ {i}^{2} \right\} = \frac{1}{2}\sum_ {i=1}^{n} \sum_ {j=1}^{n} \left(a_ {i}b_ {j} - a_ {j}b_ {i}\right)^{2} \end{equation} $
整理之后我们得到Lagrange’s Identity:
$ \begin{equation} \left(\sum_ {i=1}^{n}a_ {i}b_ {i}\right)^{2} = \sum_ {i=1}^{n}a_ {i}^{2} \sum_ {i=1}^{n}b_ {i}^{2} - \frac{1}{2}\sum_ {i=1}^{n} \sum_ {j=1}^{n}\left(a_ {i}b_ {j} - a_ {j}b_ {i}\right)^{2} \end{equation} $
EQUALITY AND A GAUGE OF PROPORTIONALITY
如果 $ \left(b_ {1}, b_ {2}, \ldots, b_ {n}\right) \ne 0 $，则存在某个 $ b_ {k} \ne 0 $，如果柯西不等式等号成立，则Lagrange’s identity右边所有项必须为0。如果我们考虑包含 $ b_ {k} $的项，有
$ \begin{equation} a_ {i}b_ {k} = a_ {k} b_ {i}, \qquad 1 \le i \le n\end{equation} $
如果我们带入 $ \lambda = \frac{a_ {k}}{b_ {k}} $，则有
$ \begin{equation} a_ {i} = \lambda b_ {i} \qquad 1 \le i \le n \end{equation} $
即Lagrange’s identity告诉我们对非零序列使得柯西不等式的等号成立当且仅当两个序列是成比例的。
现在对对称公式
$ \begin{equation} Q_ {n} = \frac{1}{2} \sum_ {i=1}^{n} \sum_ {j=1}^{n} \left(a_ {i}b_ {j} - a_ {j}b_ {i}\right)^{2} \end{equation} $
有两个有用的解释。我们引入它本来是测量柯西不等式两边的差异，现在我们看到它也能测量两个向量 $ \left(a_ {1}, a_ {2}, \ldots, a_ {n}\right)和\left(b_ {1}, b_ {2}, \ldots, b_ {n}\right) $是成比例的。
exercises
Milne and Gauges of Proportionality
如果我们限制向量中的元素为正，则可使用如下和
$ \begin{equation} R = \frac{1}{2} \sum_ {i=1}^{n} \sum_ {j=1}^{n} \frac{\left(a_ {i}b_ {j} - a_ {j}b_ {i}\right)^{2}}{\left(a_ {i} +b_ {i}\right)\left(a_ {j} + b_ {j}\right)} \end{equation} $
利用这个等式证明E.A. Milne的不等式：
$ \begin{equation} \left\{ \sum_ {j=1}^{n} \left(a_ {j} + b_ {j}\right) \right\} \left\{ \sum_ {j=1}^{n} \frac{a_ {j}b_ {j}}{\left(a_ {j} + b_ {j}\right)} \right\} \le \left\{ \sum_ {j=1}^{n} a_ {j} \right\} \left\{ \sum_ {j=1}^{n} b_ {j} \right\} \end{equation} $
并确定当且仅当向量 $ \left(a_ {1}, a_ {2}, \ldots, a_ {n} \right) 和 \left(b_ {1}, b_ {2}, \ldots, b_ {n} \right) $成比例时等号成立
证明：利用如下等式改写等式两边的形式：
$ \begin{equation} \left\{ \sum_ {j=1}^{n} a_ {j} \right\} \left\{ \sum_ {j=1}^{n} b_ {j} \right\} = \sum_ {1 \le i < j \le n} \left(a_ {i}b_ {j} + a_ {j} b_ {i} \right) \end{equation} $
然后左边减去右边即可得到R形式的式子，则得证