Forgot password
 Register account
View 8|Reply 1

[几何] Lagrange恒等式向量形式

[Copy link]

3269

Threads

7881

Posts

52

Reputation

Show all posts

hbghlyj posted 2025-8-4 10:43 |Read mode
设 $P_1, P_2, \dots, P_n$ 是单位球上的点。证明:$$\sum_{1 \leq i<j \leq n} P_iP_j^2 \leq n^2$$

3269

Threads

7881

Posts

52

Reputation

Show all posts

original poster hbghlyj posted 2025-8-4 10:48
将这些点视为单位向量 $v_1, v_2, \dots, v_n$,距离为 $P_iP_j = \|v_i - v_j\|$.

考虑Lagrange恒等式,对于任意向量 $v_1, \dots, v_n$:
$$n \left( \sum_{i=1}^n \|v_i\|^2 \right) - \left\| \sum_{i=1}^n v_i \right\|^2 = \sum_{i < j} \|v_i - v_j\|^2$$
验证
在一维情况下:对于实数 $a_1, \dots, a_n$,有$$\left(\sum 1^2 \right)\left( \sum a_i^2 \right)  - \left(\sum a_i \right)^2 = \sum_{i < j} (a_i - a_j)^2$$
我们可以将恒等式应用到每个坐标上,然后求和。
那么:
$$\sum_{i < j} \|v_i - v_j\|^2 = n \sum \|v_i\|^2 - \left\| \sum v_i \right\|^2 = n^2 - \left\| \sum v_i \right\|^2$$
因此 $\sum_{i < j} \|v_i - v_j\|^2 \leq n^2$,等号成立当且仅当 $\sum v_i=0$,质心在原点。

Quick Reply

Advanced Mode
B Color Image Link Quote Code Smilies
You have to log in before you can reply Login | Register account

$\LaTeX$ formula tutorial

Mobile version

2025-8-4 18:51 GMT+8

Powered by Discuz!

Processed in 0.017780 seconds, 43 queries