求P点位置，使表达式最小

realnumber

三角形ABC所在平面内，求动点P的位置，使得$\abs{\vv{PA}}^2+\abs{\vv{PB}}^2+\abs{\vv{PC}}^2$最小.
用直角坐标系做，P(x,y),A($x_1,y_1$),...,这个可以得到P为ABC重心
但小伙轻描淡写说用向量直接化简更简单，问题是我没明白他的办法，时候也不早了，也不明白这个办法

Tesla35

有恒等式
$\vv{PA}^2+\vv{PB}^2+\vv{PC}^2

=~(\vv{PG}+\vv{GA})^2+(\vv{PG}+\vv{GB})^2+(\vv{PG}+\vv{GC})^2$

$=~\vv{GA}^2+\vv{GB}^2+\vv{GC}^2+3\vv{PG}^2+2\vv{PG}\cdot(\vv{GA}+\vv{GB}+\vv{GC})$

$=~\vv{GA}^2+\vv{GB}^2+\vv{GC}^2+3\vv{PG}^2$.

hbghlyj

A Geometrical Theorem of Leibniz
a generalization of a leibniz geometrical theorem - arXiv
Two New Proofs of Leibniz's Inequality - Art of Problem Solving

player1703

设$x_1,x_2,\dots,x_n \in\mathbb{R}$, 当$c=\overline{x} = \frac{1}{n}\sum_\limits{i=1}^{n}{x_i}$时$\sum_\limits{i=1}^{n}{(x_i - c)^2}$最小.
容易推广到多维的情形:
设$\bm x_1,\bm x_2,\dots,\bm x_n \in\mathbb{R}^m$, 当$\bm c = \overline{\bm x} = \frac{1}{n}\sum_\limits{i=1}^{n}{\bm x_i}$ 时 $\sum_\limits{i=1}^{n}{\lVert{\bm x_i - \bm c}\rVert^2}$最小.
如果懂多元函数求导就很好证明, 令$f(\bm c) = \sum_\limits{i=1}^{n}{\lVert\bm x_i - \bm c\rVert^2}$, $\nabla f = 2\sum_\limits{i=1}^{n}{(\bm c - \bm x_i)} = \bm0$, 易得$\bm c = \frac{1}{n}\sum_\limits{i=1}^{n}{\bm x_i}$.
不用求导也很容易, 可以用类似上面解答的向量运算方法
\begin{align*}
\sum_{i=1}^{n}{\lVert\bm x_i - \bm c\rVert^2} & = \sum_{i=1}^{n}{\lVert\bm x_i - \overline{\bm x} + \overline{\bm x} - \bm c\rVert^2} \\
& = \sum_{i=1}^{n}{(\lVert\bm x_i - \overline{\bm x}\rVert^2 + 2(\bm x_i - \overline{\bm x})\cdot(\bm c - \overline{\bm x}) + \lVert\bm c - \overline{\bm x}\rVert^2}) \\
& = \sum_{i=1}^{n}{\lVert\bm x_i - \overline{\bm x}\rVert^2} + 2\sum_{i=1}^{n}{(\bm x_i - \overline{\bm x})\cdot(\bm c - \overline{\bm x})} + n\lVert\bm c - \overline{\bm x}\rVert^2
\end{align*}
其中$\sum_\limits{i=1}^{n}{(\bm x_i - \overline{\bm x})\cdot(\bm c - \overline{\bm x})} = {(\bm c - \overline{\bm x})\cdot\sum_\limits{i=1}^{n}(\bm x_i - \overline{\bm x})} = \bm 0$, 所以:
\begin{equation*}
\sum_{i=1}^{n}{\lVert\bm x_i - \bm c\rVert^2} =  \sum_{i=1}^{n}{\lVert\bm x_i - \overline{\bm x}\rVert^2} + n\lVert\bm c - \overline{\bm x}\rVert^2
\end{equation*}
显然当$\bm c = \overline{\bm x}$时取最小值$\sum_\limits{i=1}^{n}{\lVert\bm x_i - \overline{\bm x}\rVert^2}$.

更新: 其实$\bm c = (c_1,c_2,\dots,c_m)^T$把坐标拆开算简化成$m$个一独立的一维情形也行:
\begin{align*}
\sum_{i=1}^{n}{\lVert\bm x_i - \bm c\rVert^2} & = \sum_{i=1}^{n}{\sum_{j=1}^m(x_{ij}-c_j)^2} \\
&=\sum_{j=1}^{m}{\sum_{i=1}^n(x_{ij}-c_j)^2}
\end{align*}
由于$\bm c$各个分量是独立的只要使得对每一个$j=1,2,\dots,m$ ${\sum_\limits{i=1}^n(x_{ij}-c_j)^2} $最小即可所以取$c_j=\frac{1}{n}\sum_\limits{i=1}^n{x_{ij}} \xlongequal[]{记作} \overline{x_{\cdot j}}$, 合起来就是:
\begin{equation*}
\bm c = \begin{pmatrix} c_1 \\ c_2 \\ \vdots \\ c_m \end{pmatrix} = \begin{pmatrix} \overline{x_{\cdot 1}} \\ \overline{x_{\cdot 2}} \\ \vdots \\ \overline{x_{\cdot m}} \end{pmatrix} = \overline{\bm x}
\end{equation*}