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本文分为 3 部分，

第 1 部分是介绍这些不等式，第 2 部分是证明这些不等式，第 3 部分为小量习题和参考资料

可爱的不等式们

均值不等式：

柯西 (Cauchy)不等式

$$(\sum_{i=1}^na_i^2)\times (\sum_{i=1}^nb_i^2)\ge (\sum _{i=1}^na_ib_i)^2$$

排序不等式

$$\sum_{i=1}^na_ib_{n-i+1}\le \sum_{i=1}^na_ib_{d_i}\le \sum _{i=1}^n{a_ib_i}$$

Extra: （补充的可以作为了解的几个不等式）

杨氏 (Young) 不等式

$$ab\le \frac {a^p}p +\frac {b^q}q$$

赫尔德 (Holder) 不等式

$$\sum_{i=1}^n a_ib_i\le \sqrt[p]{\sum _{i=1}^na^p_i}\times \sqrt[q]{{\sum_{i=1}^n}b_i^q}$$

证明

均值不等式：

$$(c-d)^2\ge 0$$ $$c^2-2cd+d^2\ge 0$$ $$c^2+d^2\ge 2cd$$ $$\sum_{1\le i,j\le n,i\ne j} {2a_ia_j} \le (n-1)\times \sum _{i=1}^n a_i^2$$ $$\sum_{1\le i,j\le n,i\ne j} {2a_ia_j} +\sum _{i=1}^n a_i^2\le n\times \sum _{i=1}^n a_i^2$$ $$(\sum_{i=1}^na_i)^2\le n\times \sum _{i=1}^n a_i^2$$ $$\frac {(\sum_{i=1}^na_i)^2}{n^2}\le \frac {\sum _{i=1}^n a_i^2}{n}$$ $$\frac {\sum_{i=1}^na_i}{n}\le \sqrt{\frac {\sum _{i=1}^n a_i^2}{n}}$$ $$(\frac {c+d}2)^2-cd$$ $$= \frac {c^2+2cd+d^2}{4}-cd$$ $$=\frac {c^2-2cd+d^2}{4}=\frac{(c-d)^2}{4}$$ $$(c-d)^2 \ge 0$$ $$\frac{(c-d)^2}{4} \ge 0$$ $$(\frac {c+d}2)^2 \ge cd$$ $$\frac {c+d}2 \ge \sqrt{cd}$$ $$\frac {c_1+c_2}{2} \ge \sqrt{d_1d_2}$$ $$c_1 \ge d_1,c_2\ge d_2$$ $$\frac {c_1+c_2}{2}\ge \sqrt{d_1d_2}$$ $$\frac {\frac {a_1+a_2+\cdots +a_n}{n}+\frac{a_{n+1}+a_{n+2}+\cdots +a_{2n}}{n}}{2}\ge \sqrt{\sqrt[n]{a_1a_2\cdots a_n} \times \sqrt[n]{a_{n+1}a_{n+2}\cdots a_{2n}}}$$ $$\frac {\frac {a_1+a_2+...+a_{2n}}{n}}{2}\ge \sqrt{\sqrt[n]{a_1a_2...a_{2n}}}$$ $$\frac {a_1+a_2+...+a_{2n}}{2n}\ge \sqrt[2n]{a_1a_2...a_{2n}}$$ $$\frac {a_1+a_2+\cdots +a_n}{n}=\frac {a_1+a_2+\cdots +a_{n-1}}{n-1}$$ $$(n-1) \times (a_1+a_2+\cdots +a_n)=n \times (a_1+a_2+\cdots +a_{n-1})$$ $$(n-1) \times (a_1+a_2+\cdots +a_{n-1})+(n-1) \times a_n=(n-1) \times (a_1+a_2+\cdots +a_{n-1})+(a_1+\cdots +a_{n-1})$$ $$(n-1) \times a_n=(a_1+a_2+\cdots +a_{n-1})$$ $$a_n=\frac {a_1+a_2+\cdots +a_{n-1}}{n-1}$$ $$\frac {a_1+a_2+\cdots +a_{n-1}}{n-1} \ge \sqrt[n]{a_1a_2\cdots a_n}$$ $$(\frac {a_1+a_2+...+a_{n-1}}{n-1})^n \ge a_1a_2...a_n$$ $$(\frac {a_1+a_2+\cdots +a_{n-1}}{n-1})^n \ge a_1a_2\cdots a_{n-1} \times \frac {a_1+a_2+\cdots +a_{n-1}}{n-1}$$ $$(\frac {a_1+a_2+\cdots +a_{n-1}}{n-1})^{n-1} \ge a_1a_2\cdots a_{n-1}$$ $$\frac {a_1+a_2+\cdots +a_{n-1}}{n-1} \ge \sqrt[n-1]{a_1a_2\cdots a_{n-1}}$$ $$\sqrt[n]{a_1a_2\cdots a_n}\le \frac {\sum_{i=1}^na_i}n$$ $$\sqrt[n]{\frac 1{a_1a_2\cdots a_n}}\le \frac {\sum_{i=1}^n\frac 1{a_i}}n$$ $$\frac 1{\sqrt[n]{\frac 1{a_1a_2\cdots a_n}}}\ge \frac n{\sum_{i=1}^n\frac 1{a_i}}$$ $$((\frac 1{a_1a_2\cdots a_n})^{\frac 1n})^{-1}\ge \frac n{\sum_{i=1}^n\frac 1{a_i}}$$ $$((\frac 1{a_1a_2\cdots a_n})^{-1})^{\frac 1n}\ge \frac n{\sum_{i=1}^n\frac 1{a_i}}$$ $$(a_1a_2\cdots a_n)^{\frac 1n}\ge \frac n{\sum_{i=1}^n\frac 1{a_i}}$$ $$\sqrt[n]{a_1a_2\cdots a_n)}\ge \frac n{\sum_{i=1}^n\frac 1{a_i}}$$

柯西 (Cauchy)不等式

$$(\sum_{i=1}^na_i^2)\times (\sum_{i=1}^nb_i^2)\ge (\sum _{i=1}^na_ib_i)^2$$ $$f(x)=\sum_{i=1}^n(a_ix+b_i)^2 =(\sum_{i=1}^na_i^2)x^2+2(\sum_{i=1}^na_ib_i)+\sum_{i=1}^nb_i^2\ge 0$$ $$(2(\sum_{i=1}^na_ib_i))^2-4((\sum_{i=1}^na_i^2)\times (\sum_{i=1}^nb_i^2))\le 0$$ $$4(\sum_{i=1}^na_ib_i)^2\le4((\sum_{i=1}^na_i^2)\times (\sum_{i=1}^nb_i^2))$$ $$(\sum_{i=1}^na_i^2)\times (\sum_{i=1}^nb_i^2)\ge(\sum_{i=1}^na_ib_i)^2$$ $$(a^2+b^2)(c^2+d^2)=(ac)^2+(bd)^2+(bc)^2+(ad)^2$$ $$=(ac)^2+2abcd+(bd)^2+(bc)^2-2abcd+(ad)^2$$ $$=(ac+bd)^2-(bc-ad)^2\ge (ac+bd)^2$$ $$(\sum_{i=1}^na_i^2+a_{n+1}^2)\times (\sum_{i=1}^nb_i^2+b_{n+1}^2)=(\sum_{i=1}^na_i^2)\times (\sum_{i=1}^nb_i^2)+(\sum_{i=1}^na_i^2)\times b_{n+1}^2+(\sum_{i=1}^nb_i^2)\times a_{n+1}^2+a_{n+1}^2b_{n+1}^2$$ $$(AD^2-BC^2)\ge0$$ $$A^2D^4-2ABC^2D^2+B^2C^4\ge 0$$ $$A^2D^4+2ABC^2D^2+B^2C^4\ge 4ABC^2D^2$$ $$(AD^2+BC^2)^2\ge 4ABC^2D^2$$ $$AD^2+BC^2\ge \sqrt{4ABC^2D^2}$$ $$A\times D^2+B\times C^2\ge 2\times |C|\times |D|\times \sqrt{AB}$$ $$(\sum_{i=1}^na_i^2)\times b_{n+1}^2+(\sum_{i=1}^nb_i^2)\times a_{n+1}^2\ge2\times |a_{n+1}|\times |b_{n+1}|\times \sqrt{(\sum_{i=1}^na_i^2)\times (\sum_{i=1}^nb_i^2)}$$ $$(\sum_{i=1}^na_i^2)\times b_{n+1}^2+(\sum_{i=1}^nb_i^2)\times a_{n+1}^2\ge 2\times a_{n+1}\times b_{n+1}\times (\sum _{i=1}^na_ib_i)$$ $$(\sum_{i=1}^na_i^2)\times (\sum_{i=1}^nb_i^2)+(\sum_{i=1}^na_i^2)\times b_{n+1}^2+(\sum_{i=1}^nb_i^2)\times a_{n+1}^2+a_{n+1}^2b_{n+1}^2$$ $$\ge(\sum _{i=1}^na_ib_i)^2+2\times a_{n+1}\times b_{n+1}\times (\sum _{i=1}^na_ib_i)+a_{n+1}^2b_{n+1}^2$$ $$(\sum_{i=1}^na_i^2+a_{n+1}^2)\times (\sum_{i=1}^nb_i^2+b_{n+1}^2)\ge (\sum _{i=1}^{n+1}a_ib_i)^2$$

排序不等式

$$\sum_{i=1}^na_ib_{n-i+1}\le \sum_{i=1}^na_ib_{d_i}\le \sum _{i=1}^n{a_ib_i}$$ $$\sum _{i=1}^n{a_ib_i}+a_{n+1}b_{n+1}\ge\sum_{i=1}^na_ib_{d_i}+a_{n+1}b_{n+1}$$ $$\sum _{i=1}^{n+1}{a_ib_i}\ge\sum_{i=1}^na_ib_{d_i}+a_{n+1}b_{n+1}$$ $$\sum _{i=1}^n{a_ib_{n-i+1}}+a_{n+1}b_{n+1}\le\sum_{i=1}^na_ib_{d_i}+a_{n+1}b_{n+1}$$ $$\sum _{i=1}^{n+1}{a_ib_{n-i+1}}\le\sum_{i=1}^na_ib_{d_i}+a_{n+1}b_{n+1}$$

习题以及参考资料

$$\texttt{Thanks for reading}$$