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赫尔德不等式证明闵可夫斯基不等式

作者:互联网

\because (a+b)^{n} =(a+b) ^{n-1}a + (a+b)^{n-1}b

\therefore \sum_{i=1}^{n} (a_{i}+b_{i})^{n}=\sum_{i=1}^{n} (a_{i}+b_{i})^{n-1}a_{i}+\sum_{i=1}^{n} (a_{i}+b_{i})^{n-1}b_{i}

\because \frac{n-1}{n}+\frac{1}{n} = 1(n>1)

由赫尔德不等式可得:

\sum_{i=1}^{n} (a_{i}+b_{i})^{n-1}a_{i}+\sum_{i=1}^{n} (a_{i}+b_{i})^{n-1}b_{i}

\leq [\sum_{i=1}^{n}(a_{i}+b_{i})^{n}]^{\frac{n-1}{n}}[\sum_{i=1}^{n}a_{i}^{n}]^{\frac{1}{n}}+[\sum_{i=1}^{n}(a_{i}+b_{i})^{n}]^{\frac{n-1}{n}}[\sum_{i=1}^{n}b_{i}^{n}]^{\frac{1}{n}}

两边同时乘以[\sum_{i=1}^{n}(a_{i}+b_{i})^{n}]^{-\frac{n-1}{n}}得到闵可夫斯基不等式:

[\sum_{i=1}^{n} (a_{i}+b_{i})^{n}]^{\frac{1}{n}}\leq[\sum_{i=1}^{n}a_{i}^{n}]^{\frac{1}{n}}+[\sum_{i=1}^{n}b_{i}^{n}]^{\frac{1}{n}}(n\geq 1,a_{i}> 0,b_{i}> 0)

等号成立条件:\frac{a_{1}}{b_{1}}=\frac{a_{2}}{b_{2}}=...=\frac{a_{n}}{b_{n}}n=1

标签:7D%,不等式,5E%,赫尔德,5Cfrac%,7Bn%,闵可,7Bi%,_%
来源: https://blog.csdn.net/weixin_41170664/article/details/86692437