2022.10.5 若干代数题

  1. 链接\(\forall a,b,c\ge 0\)且满足\((a^2+b^2)(b^2+c^2)(c^2+a^2)=2\),求\(a+b+c\)的最值
    思考 三元换二元

  2. 链接\(a,b,c\ge 0\)\(ab+bc+ca=1\),求

\[P=\frac{a(b+c)}{a+\sqrt{bc}}+\frac{b(c+a)}{b+\sqrt{ca}}+\frac{c(a+b)}{c+\sqrt{ab}} \]

  1. 链接 给定实数\(a,b,c\in [0,1]\),证明\(\sqrt{|-a+b+c|}+\sqrt{|a-b+c|}+\sqrt{|a+b-c|}\le 3\)

  2. 链接 对非负实数\(a,b,c\)满足\(a+b+c=3\),求证

\[a\sqrt{\frac{a+1}{b+1}}+b\sqrt{\frac{b+1}{c+1}}+c\sqrt{\frac{c+1}{a+1}}\ge 3 \]

  1. 链接\(a,b,c\ge 0\)\(ab+bc+ca=3\),证明

\[\sqrt{\frac{a^2+b}{a+b}}+\sqrt{\frac{b^2+c}{b+c}}+\sqrt{\frac{c^2+a}{c+a}}\ge 3 \]

  1. 链接\(a,b,c\ge 0\)\(ab+bc+ca=3\),证明

\[\frac{\sqrt{a+3}}{a+\sqrt{bc}}+\frac{\sqrt{b+3}}{b+\sqrt{ca}}+\frac{\sqrt{c+3}}{c+\sqrt{ab}}\ge\frac{2(\sqrt{a}+\sqrt{b}+\sqrt{c})}{\sqrt{a+b+c+1}} \]

  1. \(x_0=1,x_n=\frac{x_{n-1}}{2}+\frac{1}{x_{n-1}},n\in \mathbb{N}_+\),证明\(x_{10}-\sqrt{2}<10^{-666}\)

  2. 数列\({x_n}\)定义如下:\(x_1=5,x_{n+1}=\frac{x_n^2+5x_n+4}{x_n+6},n\ge 1\),证明\(4+{(\frac{9}{10})}^{n-1}<x_n<4+\frac{9}{8}{(\frac{9}{10})}^{n-1}\)

  3. 设实数数列\(x_1,x_2,\cdots,x_{2020}\)满足一下两个条件\((1) x_1=x_{2020} (2)x_i+\frac{3}{x_i}=3x_{i+1}+\frac{1}{x_{i+1}},i=1,2,\cdots,2019\),求所有满足上述条件的序列中\(x_1\)的最大值

  4. 设数列\(x_0,x_1,x_2,\cdots,x_n\)满足\(x_0=\frac{1}{2}\),且\(x_{k+1}=x_k+\frac{1}{n}x_k^2,k=0,1,2,\cdots,n-1\),求证\(1-\frac{1}{n}<x_n<1\)

  5. 若函数\(f(x)=\frac{1}{3}x^3+ax^2+bx+c\)\([-1,2]\)上有三个零点,求\(f(-1)\cdot f(2)\)的取值范围

  6. 链接 求所有函数\(f:\mathbb{R} -> \mathbb{R}\),使得下式对于任意实数对\((x,y)\)成立:

\[f(2f(xy)+xf(y)+f(x))=3yf(x)+x \]

  1. 设整数\(n\ge 2\)\(a_1,a_2,\cdots,a_n\)为两两不同的实数,证明:$$\sum_{k=1}^n\prod_{j=1,j\ne k}^n\frac{1}{a_k-a_j}=0$$

  2. 对任意满足\(xyz+x+y+z=4\)的实数\(x,y,z\),证明

\[(yz+6)^2+(zx+6)^2+(xy+6)^2\ge8(xyz+5) \]

  1. 已知数列\({x_n}\)满足\(x_1=2\),对任意\(n\in\mathbb{Z}^+\)均有\(x_{n+1}=\sqrt{x_n+8}-\sqrt{x_n+3}\)
    (1) 证明:数列\(x_n\)收敛,并求其极限
    (2) 对于任意\(n\in\mathbb{Z}^+\),证明\(n\le x_1+x_2+\cdots+x_n\le n+1\)

  2. 设整数\(n\ge 2\),实数\(x_1,x_2,\cdots,x_n\)满足\(\sum_{i=1}^n x_i=0,\sum_{i=1}^n x_i^2=1\),试求\(\sum_{i=1}^n |x_i|\)的最值

  3. 证明:存在唯一的函数\(f(x,y)\),这里\(x,y\in\mathbb{Z}^+\),使得对任意\(x,y\in\mathbb{Z}^+\),均有

\[\begin{aligned} f(x,x)&=x\\ f(x,y)&=f(y,x)\\ (x+y)f(x,y)&=yf(x,x+y) \end{aligned} \]

  1. \(n\in\mathbb{Z}^+\),对于每一个\(i,j\in\{1,2,\cdots,n\}\)\(a_{ij}\)为正实数,满足\(a_{ij}a_{ji}=1\)。设\(c_i=\sum_{k=1}^n a_{ki}(i=1,2,\cdots,n)\),证明\(\sum_{i=1}^n\frac{1}{c_i}\le 1\)

  2. \(a_1,a_2,\cdots\)为整数列,\(d\)为整数,对任意\(n\in\mathbb{Z}^+\),数列\(\{a_n\}\)满足
    (1) \(|a_n|\)为素数 (2) \(a_{n+2}=a_{n+1}+a_n+d\)
    证明:\(\{a_n\}\)为常数列

  3. \(a\le a_i\le A,b\le b_i\le B(i=1,2,\cdots,n)\),\(a,b\)均为正数,则

\[(a_1^2+a_2^2+\cdots+a_n^2)(b_1^2+b_2^2+\cdots+b_n^2)\le(a_1b_1+a_2b_2+\cdots+a_nb_n)^2(\frac{\sqrt{\frac{AB}{ab}}+\sqrt{\frac{ab}{AB}}}{2}) \]

  1. \(n\in\mathbb{Z}^+\),实数\(x_1,x_2,\cdots,x_n\)满足\(x_1\le x_2\le \cdots\le x_n\)
    (1) 证明

\[(\sum_{i,j=1}^n|x_i-x_j|)^2\le\frac{2(n^2-1)}{3}\sum_{i,j=1}^n(x_i-x_j)^2 \]

(2) 证明:等号成立的充要条件是\(x_1,x_2,\cdots,x_n\)成等差数列

  1. \(a_1,a_2,\cdots,a_n\)是给定的不全为0的实数,且

\[r_1(x_1-a_1)+r_2(x_2-a_2)+\cdots+r_n(x_n-a_n)\le\sqrt{x_1^2+x_2^2+\cdots+x_n^2}-\sqrt{a_1^2+a_2^2+\cdots+a_n^2} \]

对一切\(x_1,x_2,\cdots,x_n\in\mathbb{R}\)成立,试求\(r_1,r_2,\cdots,r_n\)

  1. 已知正实数\(a,b,c\)满足\(a+b+c=3\),求证

\[\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\le\frac{a}{b^2}+\frac{b}{c^2}+\frac{c}{a^2} \]

posted @ 2022-10-05 15:23  毕天驰  阅读(84)  评论(0)    收藏  举报