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证明 根据定义 1，有
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\begin{align*} [\boldsymbol{x},\boldsymbol{y}]^2 & = (x_1 y_1 + x_2 y_2 + \cdots + x_n y_n)^2 \\ [\boldsymbol{x},\boldsymbol{x}] [\boldsymbol{y},\boldsymbol{y}] & = (x_1^2 + x_2^2 + \cdots + x_n^2) (y_1^2 + y_2^2 + \cdots + y_n^2) \end{align*}
[x,y]2[x,x][y,y]=(x1y1+x2y2+⋯+xnyn)2=(x12+x22+⋯+xn2)(y12+y22+⋯+yn2) 如果
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\boldsymbol{x} = \boldsymbol{0}
x=0：显然有
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[\boldsymbol{x},\boldsymbol{y}]^2 = [\boldsymbol{x},\boldsymbol{x}] [\boldsymbol{y},\boldsymbol{y}] = 0
[x,y]2=[x,x][y,y]=0，满足不等式。
如果
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\boldsymbol{x} \ne \boldsymbol{0}
x=0：因为根据性质 4，有
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[\boldsymbol{x},\boldsymbol{x}] = x_1^2 + x_2^2 + \cdots + x_n^2 > 0
[x,x]=x12+x22+⋯+xn2>0，所以，可以构造二次函数
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\begin{align*} f(t) & = (x_1^2 + x_2^2 + \cdots + x_n^2) t^2 + 2 (x_1 y_1 + x_2 y_2 + \cdots x_n y_n) t + (y_1^2 + y_2^2 + \cdots + y_n^2) \\ & = (x_1^2 t^2 + 2 x_1 y_1 t + y_1^2) + (x_2^2 t^2 + 2 x_2 y_2 t + y_2^2) + \cdots + (x_n^2 t^2 + 2 x_n y_n t + y_n^2 ) \\ & = (x_1 t + y_1)^2 + (x_2 t + y_2)^2 + \cdots + (x_n t + y_n)^2 \end{align*}
f(t)=(x12+x22+⋯+xn2)t2+2(x1y1+x2y2+⋯xnyn)t+(y12+y22+⋯+yn2)=(x12t2+2x1y1t+y12)+(x22t2+2x2y2t+y22)+⋯+(xn2t2+2xnynt+yn2)=(x1t+y1)2+(x2t+y2)2+⋯+(xnt+yn)2 因为对于任意
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f(t)≥0，所以
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f(t) = 0
f(t)=0 无解，进而
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f(t) 的判别式应该小于
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\Delta = [2 (x_1 y_1 + x_2 y_2 + \cdots x_n y_n)]^2 - 4 (x_1^2 + x_2^2 + \cdots + x_n^2) (y_1^2 + y_2^2 + \cdots + y_n^2) < 0
Δ=[2(x1y1+x2y2+⋯xnyn)]2−4(x12+x22+⋯+xn2)(y12+y22+⋯+yn2) |