[线性代数]向量2

对于所有 x , y ∈ R n , ∥ x + y ∥ ≤ ∥ x ∥ + ∥ y ∥ x, y \in \Bbb R^n, \|x+y\| \leq \|x\|+\|y\| x,y∈Rn,∥x+y∥≤∥x∥+∥y∥，其中对于 x ∈ R n x \in \Bbb R^n x∈Rn, ∥ x ∥ 2 = ∑ i = 1 n x i 2 \|x\|_2 = \sqrt{\sum_{i=1}^{n}x^2_i} ∥x∥2​=i=1∑n​xi2​ ​

柯西不等式： ( ∑ k = 1 n a k b k ) 2 ≤ ∑ k = 1 n a k 2 ∑ k = 1 n b k 2 \left(\sum_{k=1}^na_kb_k\right)^2 \leq \sum_{k=1}^na_k^2\sum_{k=1}^nb_k^2 (k=1∑n​ak​bk​)2≤k=1∑n​ak2​k=1∑n​bk2​ 利用柯西不等式，证明过程如下： ∥ x + y ∥ = ∑ i = 1 n ( x i + y i ) 2 = ∑ i = 1 n ( x i 2 + y i 2 + 2 x i y i ) \|x+y\| = \sqrt{\sum^n_{i=1}(x_i+y_i)^2} = \sqrt{\sum^n_{i=1}(x_i^2+y_i^2+2x_iy_i}) ∥x+y∥=i=1∑n​(xi​+yi​)2 ​=i=1∑n​(xi2​+yi2​+2xi​yi​ ​) ≤ ∑ i = 1 n ( x i 2 + y i 2 ) + 2 ∑ i = 1 n x i 2 ∑ i = 1 n y i 2 \leq\sqrt{\sum^n_{i=1}(x_i^2+y_i^2)+2\sqrt{\sum_{i=1}^nx_i^2\sum^n_{i=1}y_i^2}} ≤i=1∑n​(xi2​+yi2​)+2i=1∑n​xi2​i=1∑n​yi2​ ​ ​ = ∑ i = 1 n x i 2 + ∑ i = 1 n y i 2 + 2 ∑ i = 1 n x i 2 ∑ i = 1 n y i 2 =\sqrt{\sum^n_{i=1}x_i^2+\sum^n_{i=1}y_i^2+2\sqrt{\sum^n_{i=1}x_i^2}\sqrt{\sum^n_{i=1}y_i^2}} =i=1∑n​xi2​+i=1∑n​yi2​+2i=1∑n​xi2​ ​i=1∑n​yi2​ ​ ​ = ∑ i = 1 n x i 2 + ∑ i = 1 n y i 2 =\sqrt{\sum^n_{i=1}x_i^2}+\sqrt{\sum^n_{i=1}y_i^2} =i=1∑n​xi2​ ​+i=1∑n​yi2​ ​ = ∥ x ∥ + ∥ y ∥ =\|x\|+\|y\| =∥x∥+∥y∥ 证毕