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【Cauchy-Schwarz不等式】：
设 a 1 , a 2 , … , a n a_{1},a_{2},\ldots,a_{n} a1,a2,…,an及 b 1 , b 2 , … , b n b_{1},b_{2},\ldots,b_{n} b1,b2,…,bn都是复数，那么
∣ ∑ i = 1 n a i b ‾ i ∣ 2 ≤ ( ∑ i = 1 n ∣ a i ∣ 2 ) ( ∑ i = 1 n ∣ b i ∣ 2 ) \left| \sum_{i = 1}^{n}{a_{i}{\overline{b}}_{i}} \right|^{2} \leq \left( \sum_{i = 1}^{n}\left| a_{i} \right|^{2} \right)\left( \sum_{i = 1}^{n}\left| b_{i} \right|^{2} \right)
i=1∑naibi
2≤(i=1∑n∣ai∣2)(i=1∑n∣bi∣2)
【证明】
当 n = 1 n = 1 n=1显然有等号成立。当 n > 1 n > 1 n>1时下面用数学归纳法来证明。
- 当 n = 2 n = 2 n=2时，不等式成立
这是因为
( a 1 b ‾ 1 + a 2 b ‾ 2 ) ( a ‾ 1 b 1 + a ‾ 2 b 2 ) − ( a 1 a ‾ 1 + a 2 a ‾ 2 ) ( b 1 b ‾ 1 + b 2 b ‾ 2 ) \left( a_{1}{\overline{b}}_{1} + a_{2}{\overline{b}}_{2} \right)\left( {\overline{a}}_{1}b_{1} + {\overline{a}}_{2}b_{2} \right) – \left( a_{1}{\overline{a}}_{1} + a_{2}{\overline{a}}_{2} \right)\left( b_{1}{\overline{b}}_{1} + b_{2}{\overline{b}}_{2} \right) (a1b1+a2b2)(a1b1+a2b2)−(a1a1+a2a2)(b1b1+b2b2)
= ( a 1 a ‾ 1 b 1 b ‾ 1 + a 1 a ‾ 2 b ‾ 1 b 2 + a ‾ 1 a 2 b 1 b ‾ 2 + a 2 a ‾ 2 b 2 b ‾ 2 ) − ( a 1 a ‾ 1 b 1 b ‾ 1 + a 1 a ‾ 1 b 2 b ‾ 2 + a 2 a ‾ 2 b 1 b ‾ 1 + a 2 a ‾ 2 b 2 b ‾ 2 ) = \left( a_{1}{\overline{a}}_{1}b_{1}{\overline{b}}_{1} + a_{1}{\overline{a}}_{2}{\overline{b}}_{1}b_{2} + {\overline{a}}_{1}a_{2}b_{1}{\overline{b}}_{2} + a_{2}{\overline{a}}_{2}b_{2}{\overline{b}}_{2} \right) – \left( a_{1}{\overline{a}}_{1}b_{1}{\overline{b}}_{1} + a_{1}{\overline{a}}_{1}b_{2}{\overline{b}}_{2} + a_{2}{\overline{a}}_{2}b_{1}{\overline{b}}_{1} + a_{2}{\overline{a}}_{2}b_{2}{\overline{b}}_{2} \right) =(a1a1b1b1+a1a2b1b2+a1a2b1b2+a2a2b2b2)−(a1a1b1b1+a1a1b2b2+a2a2b1b1+a2a2b2b2)
= a 1 a ‾ 2 b ‾ 1 b 2 + a ‾ 1 a 2 b 1 b ‾ 2 − a 1 a ‾ 1 b 2 b ‾ 2 − a 2 a ‾ 2 b 1 b ‾ 1 = a_{1}{\overline{a}}_{2}{\overline{b}}_{1}b_{2} + {\overline{a}}_{1}a_{2}b_{1}{\overline{b}}_{2} – a_{1}{\overline{a}}_{1}b_{2}{\overline{b}}_{2} – a_{2}{\overline{a}}_{2}b_{1}{\overline{b}}_{1} =a1a2b1b2+a1a2b1b2−a1a1b2b2−a2a2b1b1
= a ‾ 2 b ‾ 1 ( a 1 b 2 − a 2 b 1 ) + a ‾ 1 b ‾ 2 ( a 2 b 1 − a 1 b 2 ) = {\overline{a}}_{2}{\overline{b}}_{1}\left( a_{1}b_{2} – a_{2}b_{1} \right) + {\overline{a}}_{1}{\overline{b}}_{2}\left( a_{2}b_{1} – a_{1}b_{2} \right) =a2b1(a1b2−a2b1)+a1b2(a2b1−a1b2)
= − ( a ‾ 1 b ‾ 2 − a ‾ 2 b ‾ 1 ) ( a 1 b 2 − a 2 b 1 ) = − ∣ a 1 b 2 − a 2 b 1 ∣ 2 ≤ 0 = – \left( {\overline{a}}_{1}{\overline{b}}_{2} – {\overline{a}}_{2}{\overline{b}}_{1} \right)\left( a_{1}b_{2} – a_{2}b_{1} \right) = – \left| a_{1}b_{2} – a_{2}b_{1} \right|^{2} \leq 0 =−(a1b2−a2b1)(a1b2−a2b1)=−∣a1b2−a2b1∣2≤0
所以
∣ ∑ i = 1 2 a i b ‾ i ∣ 2 ≤ ( ∑ i = 1 2 ∣ a i ∣ 2 ) ( ∑ i = 1 2 ∣ b i ∣ 2 ) \left| \sum_{i = 1}^{2}{a_{i}{\overline{b}}_{i}} \right|^{2} \leq \left( \sum_{i = 1}^{2}\left| a_{i} \right|^{2} \right)\left( \sum_{i = 1}^{2}\left| b_{i} \right|^{2} \right)
i=1∑2aibi
2≤(i=1∑2∣ai∣2)(i=1∑2∣bi∣2)等号成立的条件是
a 1 b 2 − a 2 b 1 = 0 a_{1}b_{2} – a_{2}b_{1} = 0 a1b2−a2b1=0
- 假定当 n = k ( k ≥ 2 ) n = k\ (k \geq 2) n=k (k≥2)时不等式成立，下面证明当 n = k + 1 n = k + 1 n=k+1时不等式也成立
∣ ∑ i = 1 k + 1 a i b ‾ i ∣ 2 − ( ∑ i = 1 k + 1 ∣ a i ∣ 2 ) ( ∑ i = 1 k + 1 ∣ b i ∣ 2 ) \left| \sum_{i = 1}^{k + 1}{a_{i}{\overline{b}}_{i}} \right|^{2} – \left( \sum_{i = 1}^{k + 1}\left| a_{i} \right|^{2} \right)\left( \sum_{i = 1}^{k + 1}\left| b_{i} \right|^{2} \right)
i=1∑k+1aibi
2−(i=1∑k+1∣ai∣2)(i=1∑k+1∣bi∣2)= ( ∑ i = 1 k + 1 a i b ‾ i ) ( ∑ i = 1 k + 1 a ‾ i b i ) − ( ∑ i = 1 k + 1 a i a ‾ i ) ( ∑ i = 1 k + 1 b i b ‾ i ) = \left( \sum_{i = 1}^{k + 1}{a_{i}{\overline{b}}_{i}} \right)\left( \sum_{i = 1}^{k + 1}{
{\overline{a}}_{i}b_{i}} \right) – \left( \sum_{i = 1}^{k + 1}{a_{i}{\overline{a}}_{i}} \right)\left( \sum_{i = 1}^{k + 1}{b_{i}{\overline{b}}_{i}} \right) =(i=1∑k+1aibi)(i=1∑k+1aibi)−(i=1∑k+1aiai)(i=1∑k+1bibi)= ( ∑ i = 1 k a i b ‾ i ) ( ∑ i = 1 k a ‾ i b i ) + a k + 1 b ‾ k + 1 ( ∑ i = 1 k a ‾ i b i ) + a ‾ k + 1 b k + 1 ( ∑ i = 1 k a i b ‾ i ) + a k + 1 a ‾ k + 1 b k + 1 b ‾ k + 1 − ( ∑ i = 1 k a i a ‾ i ) ( ∑ i = 1 k b i b ‾ i ) − a k + 1 a ‾ k + 1 ( ∑ i = 1 k b i b ‾ i ) − b k + 1 b ‾ k + 1 ( ∑ i = 1 k a i a ‾ i ) − a k + 1 a ‾ k + 1 b k + 1 b ‾ k + 1 = \left( \sum_{i = 1}^{k}{a_{i}{\overline{b}}_{i}} \right)\left( \sum_{i = 1}^{k}{
{\overline{a}}_{i}b_{i}} \right) + a_{k + 1}{\overline{b}}_{k + 1}\left( \sum_{i = 1}^{k}{
{\overline{a}}_{i}b_{i}} \right) + {\overline{a}}_{k + 1}b_{k + 1}\left( \sum_{i = 1}^{k}{a_{i}{\overline{b}}_{i}} \right) + a_{k + 1}{\overline{a}}_{k + 1}b_{k + 1}{\overline{b}}_{k + 1} – \left( \sum_{i = 1}^{k}{a_{i}{\overline{a}}_{i}} \right)\left( \sum_{i = 1}^{k}{b_{i}{\overline{b}}_{i}} \right) – a_{k + 1}{\overline{a}}_{k + 1}\left( \sum_{i = 1}^{k}{b_{i}{\overline{b}}_{i}} \right) – b_{k + 1}{\overline{b}}_{k + 1}\left( \sum_{i = 1}^{k}{a_{i}{\overline{a}}_{i}} \right) – a_{k + 1}{\overline{a}}_{k + 1}b_{k + 1}{\overline{b}}_{k + 1} =(i=1∑kaibi)(i=1∑kaibi)+ak+1bk+1(i=1∑kaibi)+ak+1bk+1(i=1∑kaibi)+ak+1ak+1bk+1bk+1−(i=1∑kaiai)(i=1∑kbibi)−ak+1ak+1(i=1∑kbibi)−bk+1bk+1(i=1∑kaiai)−ak+1ak+1bk+1bk+1因为（根据 n = k n = k n=k时的不等式）( ∑ i = 1 k a i b ‾ i ) ( ∑ i = 1 k a ‾ i b i ) − ( ∑ i = 1 k a i a ‾ i ) ( ∑ i = 1 k b i b ‾ i ) ≤ 0 \left( \sum_{i = 1}^{k}{a_{i}{\overline{b}}_{i}} \right)\left( \sum_{i = 1}^{k}{
{\overline{a}}_{i}b_{i}} \right) – \left( \sum_{i = 1}^{k}{a_{i}{\overline{a}}_{i}} \right)\left( \sum_{i = 1}^{k}{b_{i}{\overline{b}}_{i}} \right) \leq 0 (i=1∑kaibi)(i=1∑kaibi)−(i=1∑kaiai)(i=1∑kbibi)≤0而对于
a k + 1 b ‾ k + 1 ( ∑ i = 1 k a ‾ i b i ) + a ‾ k + 1 b k + 1 ( ∑ i = 1 k a i b ‾ i ) − a k + 1 a ‾ k + 1 ( ∑ i = 1 k b i b ‾ i ) − b k + 1 b ‾ k + 1 ( ∑ i = 1 k a i a ‾ i ) a_{k + 1}{\overline{b}}_{k + 1}\left( \sum_{i = 1}^{k}{
{\overline{a}}_{i}b_{i}} \right) + {\overline{a}}_{k + 1}b_{k + 1}\left( \sum_{i = 1}^{k}{a_{i}{\overline{b}}_{i}} \right) – a_{k + 1}{\overline{a}}_{k + 1}\left( \sum_{i = 1}^{k}{b_{i}{\overline{b}}_{i}} \right) – b_{k + 1}{\overline{b}}_{k + 1}\left( \sum_{i = 1}^{k}{a_{i}{\overline{a}}_{i}} \right) ak+1bk+1(i=1∑kaibi)+ak+1bk+1(i=1∑kaibi)−ak+1ak+1(i=1∑kbibi)−bk+1bk+1(i=1∑kaiai)= ∑ i = 1 k ( a k + 1 a ‾ i b ‾ k + 1 b i + a ‾ k + 1 a i b k + 1 b ‾ i − a k + 1 a ‾ k + 1 b i b ‾ i − a i a ‾ i b k + 1 b ‾ k + 1 ) = ∑ i = 1 k ( − ∣ a k + 1 b i − a i b k + 1 ∣ 2 ) ≤ 0 = \sum_{i = 1}^{k}\left( a_{k + 1}{\overline{a}}_{i}{\overline{b}}_{k + 1}b_{i} + {\overline{a}}_{k + 1}a_{i}b_{k + 1}{\overline{b}}_{i} – a_{k + 1}{\overline{a}}_{k + 1}b_{i}{\overline{b}}_{i} – a_{i}{\overline{a}}_{i}b_{k + 1}{\overline{b}}_{k + 1} \right) = \sum_{i = 1}^{k}\left( – \left| a_{k + 1}b_{i} – a_{i}b_{k + 1} \right|^{2} \right) \leq 0 =i=1∑k(ak+1aibk+1bi+ak+1aibk+1bi−ak+1ak+1bibi−aiaibk+1bk+1)=i=1∑k(−∣ak+1bi−aibk+1∣2)≤0
从而
∣ ∑ i = 1 k + 1 a i b ‾ i ∣ 2 − ( ∑ i = 1 k + 1 ∣ a i ∣ 2 ) ( ∑ i = 1 k + 1 ∣ b i ∣ 2 ) \left| \sum_{i = 1}^{k + 1}{a_{i}{\overline{b}}_{i}} \right|^{2} – \left( \sum_{i = 1}^{k + 1}\left| a_{i} \right|^{2} \right)\left( \sum_{i = 1}^{k + 1}\left| b_{i} \right|^{2} \right)
i=1∑k+1aibi
2−(i=1∑k+1∣ai∣2)(i=1∑k+1∣bi∣2)= ( ∑ i = 1 k a i b ‾ i ) ( ∑ i = 1 k a ‾ i b i ) − ( ∑ i = 1 k a i a ‾ i ) ( ∑ i = 1 k b i b ‾ i ) + a k + 1 b ‾ k + 1 ( ∑ i = 1 k a ‾ i b i ) + a ‾ k + 1 b k + 1 ( ∑ i = 1 k a i b ‾ i ) − a k + 1 a ‾ k + 1 ( ∑ i = 1 k b i b ‾ i ) − b k + 1 b ‾ k + 1 ( ∑ i = 1 k a i a ‾ i ) ≤ 0 = \left( \sum_{i = 1}^{k}{a_{i}{\overline{b}}_{i}} \right)\left( \sum_{i = 1}^{k}{
{\overline{a}}_{i}b_{i}} \right) – \left( \sum_{i = 1}^{k}{a_{i}{\overline{a}}_{i}} \right)\left( \sum_{i = 1}^{k}{b_{i}{\overline{b}}_{i}} \right) + a_{k + 1}{\overline{b}}_{k + 1}\left( \sum_{i = 1}^{k}{
{\overline{a}}_{i}b_{i}} \right) + {\overline{a}}_{k + 1}b_{k + 1}\left( \sum_{i = 1}^{k}{a_{i}{\overline{b}}_{i}} \right) – a_{k + 1}{\overline{a}}_{k + 1}\left( \sum_{i = 1}^{k}{b_{i}{\overline{b}}_{i}} \right) – b_{k + 1}{\overline{b}}_{k + 1}\left( \sum_{i = 1}^{k}{a_{i}{\overline{a}}_{i}} \right) \leq 0 =(i=1∑kaibi)(i=1∑kaibi)−(i=1∑kaiai)(i=1∑kbibi)+ak+1bk+1(i=1∑kaibi)+ak+1bk+1(i=1∑kaibi)−ak+1ak+1(i=1∑kbibi)−bk+1bk+1(i=1∑kaiai)≤0即
∣ ∑ i = 1 k + 1 a i b ‾ i ∣ 2 ≤ ( ∑ i = 1 k + 1 ∣ a i ∣ 2 ) ( ∑ i = 1 k + 1 ∣ b i ∣ 2 ) \left| \sum_{i = 1}^{k + 1}{a_{i}{\overline{b}}_{i}} \right|^{2} \leq \left( \sum_{i = 1}^{k + 1}\left| a_{i} \right|^{2} \right)\left( \sum_{i = 1}^{k + 1}\left| b_{i} \right|^{2} \right)
i=1∑k+1aibi
2≤(i=1∑k+1∣ai∣2)(i=1∑k+1∣bi∣2)等号成立的条件是，对于所有下标对 i ≠ j i \neq j i=j，都有
a j b i − a i b j = 0 a_{j}b_{i} – a_{i}b_{j} = 0 ajbi−aibj=0
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