什么是柯西(Cauchy)不等式
对于任意两组实数 a 1 , a 2 , . . . , a n ; b 1 , b 2 , . . . , b n a_1, a_2, ..., a_n; b_1, b_2, ..., b_n a1,a2,...,an;b1,b2,...,bn, 有 ∑ i = 1 n a i 2 ⋅ ∑ i = 1 n b i 2 ≥ ( ∑ i = 1 n a i b i ) 2 \displaystyle \sum_{i = 1}^n a_i^2 \cdot \sum_{i = 1}^n b_i^2 \ge (\sum_{i = 1}^n a_ib_i)^2 i=1∑nai2⋅i=1∑nbi2≥(i=1∑naibi)2, 当且仅当 a i a_i ai与 b i ( i = 1 , 2 , . . . , n ) b_i (i = 1, 2, ..., n) bi(i=1,2,...,n)对应成比例, 即 a 1 b 1 = a 2 b 2 = . . . = a n b n \displaystyle \frac{a_1}{b_1} = \frac{a_2}{b_2} = ... = \frac{a_n}{b_n} b1a1=b2a2=...=bnan时等号成立. 这个不等式成为柯西(Cauchy)不等式.
柯西(Cauchy)不等式的证明
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(a_1^2 + a_2^2 + ... + a_n^2)(b_1^2 + b_1^2 + ...+ b_n^2) \ge (a_1b_1 + a_2b_2 + ... + a_nb_n)^2
(a12+a22+...+an2)(b12+b12+...+bn2)≥(a1b1+a2b2+...+anbn)2
证明
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不等式左右两边都为
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不等式成立
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所以方程
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所以此时
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时等号成立
综上所述
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命题得证
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\begin{aligned} 证明: \: &a_1^2x^2 - 2a_1b_1x + b_1^2 = (a_1x - b_1)^2 \ge 0 \\ &a_2^2x^2 - 2a_2b_2x + b_2^2 = (a_2x - b_2)^2 \ge 0 \\ &...\\ &a_n^2x^2 - 2a_nb_nx + b_n^2 = (a_nx - b_n)^2 \ge 0 \\ &相加得: (a_1^2 + a_2^2 + ... + a_n^2)x^2 - 2(a_1b_1 + a_2b_2 + ... + a_nb_n)x + (b_a^2 + b_2 ^2 + ... + b_n^2) \ge 0 \\ &\begin{aligned} 1) \: & a_1^2 + a_2^2 + ... + a_n^2 = 0 \\ &a_1 = a_2 = ... = a_n = 0 \\ &此时 \frac{a_1}{b_1} = \frac{a_2}{b_2} = ... = \frac{a_n}{b_n} = 0 \\ &不等式左右两边都为0 \\ &不等式成立 \end{aligned} \\ &\begin{aligned} 2) \: & a_1^2 + a_2^2 + ... + a_n^2 \ne 0 \\ &因为 a_1^2 + a_2^2 + ... + a_n^2 > 0 且 (a_1^2 + a_2^2 + ... + a_n^2)x^2 - 2(a_1b_1 + a_2b_2 + ... + a_nb_n)x + (b_a^2 + b_2 ^2 + ... + b_n^2) \ge 0 \\ &所以方程 (a_1^2 + a_2^2 + ... + a_n^2)x^2 - 2(a_1b_1 + a_2b_2 + ... + a_nb_n)x + (b_a^2 + b_2 ^2 + ... + b_n^2) = 0 有2个相等实根或无实数根 \\ &所以 \Delta \le 0 \\ &所以 4(a_1b_1 + a_2b_2 + ... + a_nb_n)^2 - 4(a_1^2 + a_2^2 + ... + a_n^2)(b_1^2 + b_1^2 + ... + b_n^2) \le 0 \\ &所以(a_1^2 + a_2^2 + ... + a_n^2)(b_1^2 + b_1^2 + ... + b_n^2) \ge (a_1b_1 + a_2b_2 + ... + a_nb_n)^2 \\ &当 a_1^2x^2 - 2a_1b_1x + b_1^2 = a_2^2x^2 - 2a_2b_2x + b_2^2 = ... = a_n^2x^2 - 2a_nb_nx + b_n^2 = 0 时 \\ &即 (a_1x - b_1)^2 = (a_2x - b_2)^2 = ... = (a_nx - b_n)^2 = 0时 \\ &此时 a_1x - b_1 = a_2x - b_2 = ... = a_nx - b_n = 0 \\ &此时 \frac{a_1}{b_1} = \frac{a_2}{b_2} = ... = \frac{a_n}{b_n} = \frac{1}{x} \\ &所以方程 (a_1^2 + a_2^2 + ... + a_n^2)x^2 - 2(a_1b_1 + a_2b_2 + ... + a_nb_n)x + (b_a^2 + b_2 ^2 + ... + b_n^2) = 0 有2个相等实根\\ &所以此时 \Delta = 0 \\ &所以(a_1^2 + a_2^2 + ... + a_n^2)(b_1^2 + b_1^2 + ... + b_n^2) = (a_1b_1 + a_2b_2 + ... + a_nb_n)^2 \\ &所以 \frac{a_1}{b_1} = \frac{a_2}{b_2} = ... = \frac{a_n}{b_n} 时等号成立 \end{aligned} \\ &综上所述: 命题得证. \end{aligned}
证明:a12x2−2a1b1x+b12=(a1x−b1)2≥0a22x2−2a2b2x+b22=(a2x−b2)2≥0...an2x2−2anbnx+bn2=(anx−bn)2≥0相加得:(a12+a22+...+an2)x2−2(a1b1+a2b2+...+anbn)x+(ba2+b22+...+bn2)≥01)a12+a22+...+an2=0a1=a2=...=an=0此时b1a1=b2a2=...=bnan=0不等式左右两边都为0不等式成立2)a12+a22+...+an2=0因为a12+a22+...+an2>0且(a12+a22+...+an2)x2−2(a1b1+a2b2+...+anbn)x+(ba2+b22+...+bn2)≥0所以方程(a12+a22+...+an2)x2−2(a1b1+a2b2+...+anbn)x+(ba2+b22+...+bn2)=0有2个相等实根或无实数根所以Δ≤0所以4(a1b1+a2b2+...+anbn)2−4(a12+a22+...+an2)(b12+b12+...+bn2)≤0所以(a12+a22+...+an2)(b12+b12+...+bn2)≥(a1b1+a2b2+...+anbn)2当a12x2−2a1b1x+b12=a22x2−2a2b2x+b22=...=an2x2−2anbnx+bn2=0时即(a1x−b1)2=(a2x−b2)2=...=(anx−bn)2=0时此时a1x−b1=a2x−b2=...=anx−bn=0此时b1a1=b2a2=...=bnan=x1所以方程(a12+a22+...+an2)x2−2(a1b1+a2b2+...+anbn)x+(ba2+b22+...+bn2)=0有2个相等实根所以此时Δ=0所以(a12+a22+...+an2)(b12+b12+...+bn2)=(a1b1+a2b2+...+anbn)2所以b1a1=b2a2=...=bnan时等号成立综上所述:命题得证.
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