这里对分析学基础理论, 极限和无穷级数部分做一个review\textbf{review}
knuth\text{knuth} 的具体数学中, 很多理论都是建立在分析学基础上的

数列极限的补充问题

从每一个实数列中都可以选出一个收敛的或者是趋于无穷的子列
proof\textbf{proof}\quad 其中, 收敛很好证明
01

无穷性的证明,kN,choose nkN\forall k \in \mathbb{N}, \text{choose} \ n_k \in \mathbb{N}
xnk>k and nk<nk+1,xnk|x_{n_k}| > k \ \textbf{and} \ n_k < n_{k+1}, {x_{n_k}} 构成无穷子列

example1\textbf{example1}

limk(1)kk=limninfkn(1)kk={(1)2m+1(2m+1)=1nn=2m+1 (1)2m(1)2m+1=1n+1n=2m}=0\mathop{\underline{\lim}}\limits_{k \rightarrow \infty} \frac{(-1)^{k}}{k}=\lim _{n \rightarrow \infty} \inf _{k \geqslant n} \frac{(-1)^{k}}{k}=\left\{\begin{gathered} \frac{(-1)^{2m +1}}{(2 m+1)}=-\frac{1}{n} \quad n=2 m+1 \\ \ \\ \frac{(-1)^{2 m} \cdot(-1)}{2 m+1}=\frac{-1}{n+1} \quad n=2m \end{gathered} \right\} =0

limk(1)kk=limnsupkn(1)kk={(1)2m2m=1nn=2m (1)2m+22m+2=1n+1n=2m+1}=0\mathop{\overline{\lim}}\limits _{k \rightarrow \infty} \frac{(-1)^{k}}{k}=\lim _{n \rightarrow \infty} \sup _{k \geqslant n} \frac{(-1)^{k}}{k}=\left\{\begin{gathered} \frac{(-1)^{2m}}{2 m}=\frac{1}{n} & n=2 m \\ \ \\ \frac{(-1)^{2 m+2}}{2 m+2}=\frac{1}{n+1} & n=2 m+1 \end{gathered}\right\} = 0

下极限证明语言
ε>0,iε<in=infknxkxk\forall \varepsilon>0, \quad i-\varepsilon<i_n=\inf\limits_{k \geqslant n} x_{k} \leqslant x_{k}
因为ε\varepsilon 是任意的, 所以对于xk{x_k}
任何部分极限都不能小于iεi - \varepsilon, 也就是不能小于ii

补充阅读:《不等式的秘密》

阅读这部分内容, 主要是为了提升数感, 让自己的代数变形能力有进一步的提升

AM-GM不等式

problem1\textbf{problem1}
a,b,c>0,abc=1,证明a, b, c>0, abc = 1, \text{证明}

(a+b)(b+c)(c+a)(a+1)(b+1)(c+1)ab(a+b)+bc(b+c)+ca(c+a)a+b+c+ab+bc+ca\begin{gathered} (a+b)(b+c)(c+a) \geqslant (a+1)(b+1)(c+1) \\ \Longrightarrow ab(a+b)+bc(b+c)+ca(c+a) \geqslant a+b+c+ab+bc+ca \end{gathered}

重点在于轮换对称的构造

LHScyc={a2b+a2cb2a+b2cc2a+c2b}{a2b+ab2b2c+bc2c2a+ca2}L H S_{c y c}=\left\{\begin{gathered} a^{2} b+a^{2} c \\ b^{2} a+b^{2} c \\ c^2a+c^{2} b \end{gathered}\right\} \quad \left\{\begin{gathered} a^{2} b+a b^{2} \\ b^{2} c+b c^{2} \\ c^{2} a+c a^{2} \end{gathered}\right\}

2LHS+cycab=cyc(a2b+a2b+a2c+a2c+bc)5cyca2LHS+cyca=cyc(a2b+a2b+ab2+ab2+c)5cycab\begin{gathered} 2 L H S+\sum_{c y c} a b=\sum_{c y c}\left(a^{2} b+a^{2} b+a^{2} c+a^{2} c+b c\right) \geqslant 5 \sum_{c y c}a \\ 2 L H S+\sum_{c y c} a=\sum_{c y c}\left(a^{2} b+a^{2} b+a b^{2}+a b^{2}+c\right) \geqslant 5 \sum_{c y c} a b \end{gathered}

综上, LHScyca+cycab\begin{gathered} \text{综上, }L H S \geqslant \sum_{c y c} a+\sum_{c yc} a b \end{gathered}

problem2 4n+3\textbf{problem2} \ 4n+3 是数论中非常常见的数, 因此构造34\frac{3}{4} 也是很常见的
x,y,z>0,and xyz=1x,y,z>0, \textbf{and}\ xyz=1

x3(1+y)(1+z)+y3(1+z)(1+x)+z3(1+x)(1+y)34\frac{x^{3}}{(1+y)(1+z)}+\frac{y^{3}}{(1+z)(1+x)}+\frac{z^{3}}{(1+x)(1+y)} \geq \frac{3}{4}

x3(1+y)(1+z)+1+y8+1+z834x,写成轮换对称\frac{x^{3}}{(1+y)(1+z)}+\frac{1+y}{8}+\frac{1+z}{8} \geqslant \frac{3}{4} x , \text{写成轮换对称}

cycx3(1+y)(1+z)+14cyc(1+x)cyc3x4cycx3(1+y)(1+z)14cyc(2x1)=14(x+x+y+y+)346434=34\begin{gathered} \sum_{c y c} \frac{x^{3}}{(1+y)(1+z)}+\frac{1}{4} \sum_{c y c}(1+x) \geqslant \sum_{c y c} \frac{3 x}{4} \\ \Rightarrow \sum_{c y c} \frac{x^{3}}{(1+y)(1+z)} \geq \frac{1}{4} \sum_{c y c}(2 x-1)=\frac{1}{4}(x+x+y+y+\ldots)-\frac{3}{4} \\ \geqslant \frac{6}{4}-\frac{3}{4}=\frac{3}{4} \end{gathered}

problem3轮换恒等变形\textbf{problem3轮换恒等变形}

(a2+b2+c2)(ab+bc+ca)=ab(a2+b2)+bc(b2+c2)+ca(a2+c2)+abc(a+b+c)if a,b,c0,a+b+c=2(a2+b2+c2)(ab+bc+ca)ab(a2+b2)+bc(b2+c2)+ca(a2+c2)2(a2b2+b2c2+a2c2)\begin{gathered} \left(a^{2}+b^{2}+c^{2}\right)(a b+b c+c a)=a b\left(a^{2}+b^{2}\right)+b c\left(b^{2}+c^{2}\right)+c a\left(a^{2}+c^{2}\right)+a b c(a+b+c) \\ \text{if} \ a , b, c \geqslant 0, a+b+c =2 \\ \left(a^{2}+b^{2}+c^{2}\right)(a b+b c+c a) \geqslant a b\left(a^{2}+b^{2}\right)+b c\left(b^{2}+c^{2}\right)+c a\left(a^{2}+c^{2}\right) \\ \geqslant 2\left(a^{2} b^{2}+b^2 c^{2}+a^{2} c^{2}\right) \end{gathered}

a2+b2+c2+2(ab+bc+ca)=(a+b+c)2=42(a2+b2+c2)(ab+bc+ca)(a2+b2+c2+2(ab+bc+ca)2)2=4(a2+b2+c2)(ab+bc+ca)2a2b2+b2c2+a2c21\begin{gathered} a^{2}+b^{2}+c^{2}+2(a b+b c+c a)=(a+b+c)^{2}=4 \\ 2\left(a^{2}+b^{2}+c^{2}\right)(a b+b c+c a) \leqslant\left(\frac{a^{2}+b^{2}+c^{2}+2(a b+b c+c a)}{2}\right)^{2}=4 \\ \begin{gathered} \left(a^{2}+b^{2}+c^{2}\right)(a b+b c+c a) \leqslant 2 \\ \Rightarrow a^{2} b^{2}+b^{2} c^{2}+a^{2} c^{2} \leqslant 1 \end{gathered} \end{gathered}

problem4不等式中的域分组\textbf{problem4不等式中的域分组}
a,b,c,d>0a,b,c,d>0

1a2+ab+1b2+bc+1c2+cd+1d2+da4ac+bd\frac{1}{a^{2}+a b}+\frac{1}{b^{2}+b c}+\frac{1}{c^{2}+c d}+\frac{1}{d^{2}+d a} \geqslant \frac{4}{a c+b d}

ac+bda2+ab=a2+ab+ac+bda2+ab1=a+ca+b+b(d+a)a(a+b)1LHS=cyca+ca+b+cycb(d+a)a(a+b)4\begin{gathered} \frac{a c+b d}{a^{2}+a b}=\frac{a^{2}+a b+a c+b d}{a^{2}+a b}-1=\frac{a+c}{a+b}+\frac{b(d+a)}{a(a+b)}-1 \\ L H S=\sum_{c y c} \frac{a+c}{a+b}+\sum_{c y c} \frac{b(d+a)}{a(a+b)}-4 \end{gathered}

{a+ca+b,b(d+a)a(a+b)},{b+db+c,c(a+b)b(b+c)}{c+ac+d,d(c+b)c(c+d)},{d+bd+a,a(d+c)d(d+a)} 用AM-GM不等式消去第二个项, 得到  LHScyca+ca+b\begin{gathered} \left\{\frac{a+c}{a+b}, \frac{b(d+a)}{a(a+b)}\right\}, \quad \left\{\frac{b+d}{b+c}, \frac{c(a+b)}{b(b+c)} \right\} \\ \left\{\frac{c+a}{c+d}, \frac{d(c+b)}{c(c+d)}\right\}, \quad \left\{\frac{d+b}{d+a}, \frac{a(d+c)}{d(d+a)}\right\} \\ \ \\ \text{用AM-GM不等式消去第二个项, 得到 }\\ \ \\ LH S \geqslant \sum_{c y c} \frac{a+c}{a+b} \end{gathered}

cyca+ca+b=(a+c)(1a+b+1c+d)+(b+d)(1b+c+1d+a)Schwarz 不等式 1a+b+1c+d(1+1)2a+b+c+d=4a+b+c+d LHS4(a+c)a+b+c+d+4(b+d)a+b+c+d=4\begin{gathered} \sum_{c y c} \frac{a+c}{a+b}=(a+c)\left(\frac{1}{a+b}+\frac{1}{c+d}\right)+(b+d)\left(\frac{1}{b+c}+\frac{1}{d+a}\right) \\ \text{Schwarz 不等式} \\ \ \\ \frac{1}{a+b}+\frac{1}{c+d} \geqslant \frac{(1+1)^{2}}{a+b+c+d}=\frac{4}{a+b+c+d} \\ \ \\ L H S \geqslant \frac{4(a+c)}{a+b+c+d}+\frac{4(b+d)}{a+b+c+d}=4 \end{gathered}

problem5\textbf{problem5}
a,b,c,d,e0,a+b+c+d+e=5,proofa,b,c,d,e \geqslant 0, a+b+c+d+e = 5, \text{proof}
abc+bcd+cde+dea+eab5abc+bcd+cde+dea+eab \leqslant 5
proof\textbf{proof}

abc+bcd+cde+dea+eab=e(a+c)(b+d)+bc(a+de)e(a+b+c+d2)2+(b+c+a+de3)3 =e(5e)24+(52e)227\begin{gathered} a b c+b c d+c d e+d e a+e a b=e(a+c)(b+d)+b c(a+d-e) \\ \leqslant e\left(\frac{a+b+c+d}{2}\right)^{2}+\left(\frac{b+c+a+d-e}{3}\right)^{3} \\ \ \\ =\frac{e(5-e)^{2}}{4}+\frac{(5-2 e)^{2}}{27} \end{gathered}

f(x)=x(5x)24+(52x)227df(x)dx=536(x2+4x5)x=1,df(x)dx=0 let e=min{a,b,c,d,e},e1 f(x)f(1)=25\begin{gathered} f(x)=\frac{x(5-x)^{2}}{4}+\frac{(5-2 x)^{2}}{27} \\ \frac{d f(x)}{d x}=-\frac{5}{36}\left(x^{2}+4 x-5\right) \Rightarrow x=1, \frac{d f(x)}{d x}=0 \\ \ \\ \text{let } e = \min\{a,b,c,d,e\}, \Rightarrow e \leqslant 1 \\ \ \\ f(x) \leqslant f(1)=2 \leqslant 5 \end{gathered}

problem5\textbf{problem5}
a,b,c,d>0,proofa,b,c,d>0, \text{proof}

(1a+1b+1c+1d)21a2+4a2+b2+9a2+b2+c2+16a2+b2+c2+d2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right)^{2} \geqslant \frac{1}{a^{2}}+\frac{4}{a^{2}+b^{2}}+\frac{9}{a^{2}+b^{2}+c^{2}}+\frac{16}{a^{2}+b^{2}+c^{2}+d^{2}}

1b2+1c2+1d2+sym2ab4a2+b2+9a2+b2+c2+16a2+b2+c2+d2\Leftrightarrow \frac{1}{b^{2}}+\frac{1}{c^{2}}+\frac{1}{d^{2}}+\sum_{s y m} \frac{2}{a b} \geqslant \frac{4}{a^{2}+b^{2}}+\frac{9}{a^{2}+b^{2}+c^{2}}+\frac{16}{a^{2}+b^{2}+c^{2}+d^{2}}

AM-GM不等式, 有 2ab4a2+b2 2ac+2bc(22)2ac+bc8a2+b22+c2>8a2+b2+c2 (Cauchy-Schwarz) 1b2+1c24b2+c21b2+c2+a2 d=min{a,b,c,d} 2ad+2bd+2cd(32)2ad+bd+cd 18ad+bd+cd+d218a2+b2+c2+d2>16a2+b2+c2+d2\begin{gathered} \text{AM-GM不等式, 有} \\ \ \\ \frac{2}{a b} \geqslant \frac{4}{a^{2}+b^{2}} \\ \ \\ \frac{2}{a c}+\frac{2}{b c} \geqslant \frac{(2 \sqrt{2})^{2}}{a c+b c} \geqslant \frac{8}{\frac{a^{2}+b^{2}}{2}+c^{2}}>\frac{8}{a^{2}+b^{2}+c^{2}} \ \text{(Cauchy-Schwarz)}\\ \ \\ \frac{1}{b^{2}}+\frac{1}{c^{2}} \geqslant \frac{4}{b^{2}+c^{2}} \geqslant \frac{1}{b^{2}+c^{2}+a^{2}} \\ \ \\ d=\min \{a, b, c, d\} \\ \ \\ \frac{2}{a d}+\frac{2}{b d}+\frac{2}{c d} \geqslant \frac{(3 \sqrt{2})^{2}}{a d+b d+c d} \\ \ \\ \geqslant \frac{18}{a d+b d+c d+d^{2}} \geqslant \frac{18}{a^{2}+b^{2}+c^{2}+d^{2}}>\frac{16}{a^{2}+b^{2}+c^{2}+d^{2}} \end{gathered}

problem6 askminM\textbf{problem6} \ \text{ask} \min {M}

ab(a2b2)+bc(b2c2)+ca(c2a2)M(a2+b2+c2)2\left|a b\left(a^{2}-b^{2}\right)+b c\left(b^{2}-c^{2}\right)+c a\left(c^{2}-a^{2}\right)\right| \leqslant M\left(a^{2}+b^{2}+c^{2}\right)^{2}

LHS=ab(a2b2+c2c2)+bc(b2c2)+ca(c2a2) =b(b2c2)(ca)+a(ca)(c+a)(cb) =(ab)(bc)(ca)(a+b+c) let x=ab,y=bc,z=ca,s=a+b+c\begin{gathered} L H S=a b\left(a^{2}-b^{2}+c^{2}-c^{2}\right)+b c\left(b^{2}-c^{2}\right)+c a\left(c^{2}-a^{2}\right) \\ \ \\ =b\left(b^{2}-c^{2}\right)(c-a)+a(c-a)(c+a)(c-b) \\ \ \\ =-(a-b)(b-c)(c-a)(a+b+c) \\ \ \\ \text{let } x=a-b, y = b-c, z=c-a, s=a+b+c \end{gathered}

9sxyzM(s2+x2+y2+z2)2,x+y+z=0 \begin{gathered} \Leftrightarrow 9|sxyz| \leqslant M\left(s^{2}+x^{2}+y^{2}+z^{2}\right)^{2}, \quad x +y+z =0 \\ \ \\ \end{gathered}

i)\textbf{i)}

x2+y2+z2=x2+y2+(x+y)2=2((x+y)2xy) 2((x+y)2(x+y)24)=32(x+y)2 sxyz=sxy(x+y)s(x+y)34\begin{gathered} x^{2}+y^{2}+z^{2}=x^{2}+y^{2}+(x+y)^{2}=2\left((x+y)^{2}-x y\right) \\ \ \\ \geqslant 2\left((x+y)^{2}-\frac{(x+y)^{2}}{4}\right)=\frac{3}{2}(x+y)^{2} \\ \ \\ |s x y z|=|s x y(x+y)| \leqslant |s| \frac{(x+y)^{3}}{4} \end{gathered}

ii)\textbf{ii)}

let t=x+y askst3?(s2+32t2)2 asks2t6 ?(2s2+3t2)44 we have2s2t6=2s2t2t2t2(2s2+3t24)4 42st3(2s2+3t2)24 sxyz1162(s2+x2+y2+z2)2,M=9162 {2(a+b+c)=caa+b+c=3ca=32ab=bc{b=1ca=32a+c=2 (a,b,c)=(132,1,1+32)\begin{gathered} \text{let } t = x+y \\ \ \\ \Leftrightarrow \text{ask} \quad |s| t^{3} \leqslant ?\left(s^{2}+\frac{3}{2} t^{2}\right)^{2} \\ \ \\ \Leftrightarrow \text{ask} \quad s^{2} t^{6} \leqslant \ ? \frac{\left(2 s^{2}+3 t^{2}\right)^{4}}{4} \\ \ \\ \text{we have} \quad 2 s^{2} t^{6}=2 s^{2} \cdot t^{2} \cdot t^{2} \cdot t^{2} \leqslant\left(\frac{2 s^{2}+3 t^{2}}{4}\right)^{4} \\ \ \\ 4 \cdot \sqrt{2}|s| t^{3} \leqslant \frac{\left(2 s^{2}+3 t^{2}\right)^{2}}{4} \\ \ \\ |s x y z| \leqslant \frac{1}{16 \sqrt{2}}\left(s^{2}+x^{2}+y^{2}+z^{2}\right)^{2}, M=\frac{9}{16 \sqrt{2}} \\ \ \\ \left\{\begin{gathered} \sqrt{2}(a+b+c)=c-a \\ a+b+c=3 \\ c-a=3 \sqrt{2} \\ a-b=b-c \end{gathered}\right. \Rightarrow \left\{\begin{gathered} b=1 \\ c-a=3 \sqrt{2} \\ a+c=2 \end{gathered}\right. \\ \ \\ (a, b, c)=\left(1-\frac{3}{\sqrt{2}}, 1,1+\frac{3}{\sqrt{2}}\right) \end{gathered}

Cauchy求反技术

已知a+b+c+d=4a+b+c+d = 4, 证明

a1+b2c+b1+c2d+c1+d2a+d1+a2b2\frac{a}{1+b^{2} c}+\frac{b}{1+c^{2} d}+\frac{c}{1+d^{2} a}+\frac{d}{1+a^{2} b} \geq 2

a1+b2c=aab2c1+b2ca2b2c2bcab(a+ac)4 \begin{gathered} \frac{a}{1+b^{2} c}=a-\frac{a b^{2} c}{1+b^{2} c} \geqslant a-\frac{2 b^{2} c}{2 b \sqrt{c}} \geqslant a-\frac{b(a+a c)}{4}\\ \ \\ \end{gathered}

2cycab=(cyca)2cyc(a2) (cyca)2=2cycab+cyca24cycab cycab14(cyca)2 (cyc4a)3=4cyc4a3+(31)(41)cyc4abc 16cycabc(cyca)3\begin{gathered} 2 \sum_{c y c} a b=\left(\sum_{c y c} a\right)^{2}-\sum_{c y c}\left(a^{2}\right) \Rightarrow \\ \ \\ \Rightarrow \quad\left(\sum_{cyc} a\right)^{2}=2 \sum_{c y c} a b+\sum_{c y c} a^{2} \geqslant 4 \sum_{cyc}ab \\ \ \\ \Rightarrow \sum_{c y c} a b \leqslant \frac{1}{4}\left(\sum_{c y c} a\right)^{2} \\ \ \\ \left(\sum_{c yc4} a\right)^{3}=4 \sum_{c y c 4} a^{3}+\left(\begin{gathered} 3 \\ 1 \end{gathered}\right)\left(\begin{gathered} 4 \\ 1 \end{gathered}\right) \sum_{c y c 4} a b c \\ \ \\ 16 \sum_{c y c} a b c \leqslant(\sum_{cyc} a)^{3} \end{gathered}

AM-GM and Chebyshev\textbf{AM-GM} \ \textbf{and} \ \textbf{Chebyshev}
Chebyshev\textbf{Chebyshev}

a1b1+a2b1++anbn1n(a1+a2++an)(b1+b2++bn)a_{1} b_{1}+a_{2} b_{1}+\ldots+a_{n} b_{n} \geqslant \frac{1}{n}\left(a_{1}+a_{2}+\dots+a_{n}\right)\left(b_{1}+b_{2}+\ldots+b_{n}\right)

a+b+c=3,proofa+b+c=3, \quad \text{proof}

cyc(ab)233 (a+b+c)219(a+b+c)(1+1+1)=13(a+b+c) (a+b+c)2=cyca2+2cycab3cycab (cyca)2=23(cyca)2+13(cyca)22cyca+cycab =cyc(a+a+ab)=cyc(a+b+ab)3cyc(ab)2/3\begin{gathered} \sum_{c y c}(a b)^{\frac{2}{3}} \leqslant 3 \\ \ \\ (a+b+c)^{2} \geqslant \frac{1}{9}(a+b+c)(1+1+1)=\frac{1}{3}(a+b+c) \\ \ \\ (a+b+c)^{2}=\sum_{c y c} a^{2}+2 \sum_{c y c} a b \geqslant 3 \sum_{c y c} a b \\ \ \\ \Rightarrow \left(\sum_{c yc} a\right)^{2}=\frac{2}{3}\left(\sum_{c y c} a\right)^{2}+\frac{1}{3}\left(\sum_{c y c} a\right)^{2} \geqslant 2 \sum_{c yc} a+\sum_{c y c} a b \\ \ \\ =\sum_{c y c}(a+a+a b)=\sum_{c y c}(a+b+a b) \geqslant 3 \sum_{c y c}(a b)^{2 / 3} \end{gathered}

Rearrangement inequality\textbf{Rearrangement} \ \textbf{inequality}

a+b+c=3,cycacycab cyca=13(cyca)(cyca)=13(cyca)2 a2+b2+c2ab+bc+cacyca2cycabcyca133cycab=cycab\begin{gathered} a+b+c =3, \quad \sum_{c y c} a \geqslant \sum_{c y c} a b \\ \ \\ \sum_{c y c} a=\frac{1}{3} \cdot\left(\sum_{c y c} a\right)\left(\sum_{c y c} a\right)=\frac{1}{3}\left(\sum_{c y c} a\right)^{2} \\ \ \\ \begin{gathered} &a^{2}+b^{2}+c^{2} \geqslant a b+b c+c a\\ &\sum_{c y c} a^{2} \geqslant \sum_{c y c} a b \end{gathered} \\ \sum_{c y c} a \geqslant \frac{1}{3} \cdot 3 \sum_{c y c} a b=\sum_{c y c} a b \end{gathered}

problem\textbf{problem}
a+b+c+d=4, proofa+b+c+d=4, \ \text{proof}

cycabcycab2c ab+bc+cd+daab2c+bc2d+cd2a+da2b LHS=14(b+c+d+a)(ab+bc+cd+da) ab2+bc2+cd2+da2=14(c+d+a+b)(ab2+bc2+cd2+da2) ab2c+bc2d+cd2a+da2b\begin{gathered} \sum_{c y c} a b-\sum_{c y c} a b^{2} c \\ \ \\ \Leftrightarrow a b+b c+c d+d a \leqslant a b^{2} c+b c^{2} d+c d^{2} a+d a^{2} b \\ \ \\ L H S=\frac{1}{4}(b+c+d+a)(a b+b c+c d+d a) \\ \ \\ \leqslant a b^{2}+b c^{2}+c d^{2}+d a^{2}=\frac{1}{4}(c+d+a+b)\left(a b^{2}+b c^{2}+c d^{2}+d a^{2}\right) \\ \ \\ \leqslant a b^{2} c+b c^{2} d+c d^{2} a+d a^{2} b \end{gathered}

Cauchy求反技术的技巧\textbf{Cauchy求反技术的技巧}

a,b,c>0,a2+b2+c2=3 proof 1a3+2+1b3+2+1c3+21 1a3+2=12a32(a3+2)\begin{gathered} a, b, c>0, \quad a^{2}+b^{2}+c^{2}=3 \ \text{proof} \\ \ \\ \frac{1}{a^{3}+2}+\frac{1}{b^{3}+2}+\frac{1}{c^{3}+2} \geqslant 1 \\ \ \\ \frac{1}{a^{3}+2}=\frac{1}{2}-\frac{a^{3}}{2\left(a^{3}+2\right)} \end{gathered}

连分数

连分数相关定义

表示\textbf{表示}

[a0,a1]=a0+1a1 [a0,a1,,an1,an]=[a0,a1,,an2,an1+1an]\begin{gathered} \left[a_{0}, a_{1}\right]=a_{0}+\frac{1}{a_{1}} \\ \ \\ \left[a_{0}, a_{1}, \ldots, a_{n-1}, a_{n}\right]=\left[a_{0}, a_{1}, \ldots, a_{n-2}, a_{n-1}+\frac{1}{a_{n}}\right] \end{gathered}

p0=a0,p1=a1a0+1,pn=anpn1+pn2(2nN)q0=1,q1=a1,qn=anqn1+qn2(2nN) [a0,a1,..,an]=pnqn\begin{gathered} p_{0}=a_{0}, \quad p_{1}=a_{1} a_{0}+1, \quad p_{n}=a_{n} p_{n-1}+p_{n-2} \quad(2 \leqslant n \leqslant N) \\ q_{0}=1, \quad q_{1}=a_{1}, \quad q_{n}=a_{n} q_{n-1}+q_{n-2} \quad \quad \quad(2 \leqslant n \leqslant N) \\ \ \\ \left[a_{0}, a_{1}, \ldots . ., a_{n}\right]=\frac{p_{n}}{q_{n}} \end{gathered}

proof\textbf{proof}

[a0,a1,,am1,am]=pmqm=ampm1+pm2amqm1+qm2 [a0,a1,,am1,am,am+1]=[a0,a1,.,am1,am+1am+1] =(am+1am+1)pm1+pm2(am+1am+1)qm1+qm2=am+1(ampm1+pm2)+pm1am+1(amqm1+qm2)+qm1 =pm+1qm+1\begin{gathered} \left[a_{0}, a_{1}, \ldots, a_{m-1}, a_{m}\right]=\frac{p_{m}}{q_{m}}=\frac{a_{m} p_{m-1}+p_{m-2}}{a_{m} q_{m-1}+q_{m-2}} \\ \ \\ \left[a_{0}, a_{1}, \ldots, a_{m-1}, a_{m}, a_{m+1}\right]=\left[a_{0}, a_{1}, \ldots ., a_{m-1}, a_{m}+\frac{1}{a_{m+1}}\right] \\ \ \\ =\frac{\left(a_{m}+\frac{1}{a_{m+1}}\right) p_{m-1}+p_{m-2}}{\left(a_{m}+\frac{1}{a_{m+1}}\right) q_{m-1}+q_{m-2}}=\frac{a_{m+1}\left(a_{m} p_{m-1}+p_{m-2}\right)+p_{m-1}}{a_{m+1}\left(a_{m} q_{m-1}+q_{m-2}\right)+q_{m-1}} \\ \ \\ = \frac{p_{m+1}}{q_{m+1}} \end{gathered}

lemma1\textbf{lemma1}

pnqn=anpn1+pn2anqn1+qn2\frac{p_{n}}{q_{n}}=\frac{a_{n} p_{n-1}+p_{n-2}}{a_{n} q_{n-1}+q_{n-2}}

构造递推式如下

pnqn1pn1qn=(anpn1+pn2)qn1pn1(anqn1+qn2) =(1)1(pn1qn2pn2qn1)=(1)n1(p1q0p0q1)=(1)n1\begin{gathered} p_{n} q_{n-1}-p_{n-1} q_{n}=\left(a_{n} p_{n-1}+p_{n-2}\right) q_{n-1}-p_{n-1}\left(a_{n} q_{n-1}+q_{n-2}\right) \\ \ \\ =(-1)^{1}\left(p_{n-1} q_{n-2}-p_{n-2} q_{n-1}\right)=(-1)^{n-1}\left(p_1q_{0}-p_{0} q_{1}\right)=(-1)^{n-1} \end{gathered}

还有

pnqn2pn2qn=(anpn1+pn2)qn2pn2(anqn1+qn2) =an(pn1qn2pn2qn1)=(1)n2an=(1)nan\begin{gathered} p_{n} q_{n-2}-p_{n-2} q_{n}=\left(a_{n} p_{n-1}+p_{n-2}\right) q_{n-2}-p_{n-2}\left(a_{n} q_{n-1}+q_{n-2}\right) \\ \ \\ =a_{n}\left(p_{n-1} q_{n-2}-p_{n-2} q_{n-1}\right)=(-1)^{n-2} \cdot a_{n}=(-1)^{n} a_{n} \end{gathered}

pnqn1pn1qn=(1)n1pnqn2pn2qn=(1)nan pnqnpn1qn1=(1)n1qn1qn pnqnpn2qn2=(1)nanqn2qn\begin{gathered} p_{n} q_{n-1}-p_{n-1} q_{n}=(-1)^{n-1} \\ \left.p_{n} q_{n-2}-p_{n-2} q_{n}=(-1\right)^{n} a_{n} \\ \ \\ \frac{p_{n}}{q_{n}}-\frac{p_{n-1}}{q_{n-1}}=\frac{(-1)^{n-1}}{q_{n-1} q_{n}} \\ \ \\ \frac{p_{n}}{q_{n}}-\frac{p_{n-2}}{q_{n-2}}=\frac{(-1)^{n} a_{n}}{q_{n-2} q_{n}} \end{gathered}

由此可以知道分奇偶项, 得到的单调性

lemma2\textbf{lemma2}
ifn>3,qnn\text{if} \quad n > 3, \quad q_n \geqslant n

q0=1,q1=a11qn=anqn1+qn2qn1+1=n1+1=n\begin{gathered} q_{0}=1, \quad q_{1}=a_{1} \geqslant 1 \\ q_{n}=a_{n} q_{n-1}+q_{n-2} \geqslant q_{n-1}+1=n-1+1=n \end{gathered}

根据数学归纳法, 证毕

theorem1\textbf{theorem1}
x 可以用一个奇数个渐进分数的连分数表示出来x\ \text{可以用一个奇数个渐进分数的连分数表示出来}
那么也一定能用偶数个渐进分数的连分数表示\text{那么也一定能用偶数个渐进分数的连分数表示}

[a0,a1,an]=[a0,a1,an1,1],an2 ifan=1,[a0,a1,,an1,1]=[a0,a1,,an2,an1+1]\begin{gathered} \left[a_{0}, a_{1}, \dots a_{n}\right]=\left[a_{0}, a_{1}, \dots a_{n}-1,1\right], \quad a_n \geqslant 2 \\ \ \\ \text{if} \quad a_n = 1, \left[a_{0}, a_{1}, \ldots, a_{n-1}, 1\right]=\left[a_{0}, a_{1}, \ldots, a_{n-2}, a_{n-1}+1\right] \end{gathered}

theorem2\textbf{theorem2}
连分数的第nn 个完全商
an=[an,an+1,,aN](0nN)a_{n}^{\prime}=\left[a_{n}, a_{n+1}, \ldots, a_{N}\right] \quad(0 \leqslant n \leqslant N)

x=a0,x=a0+1a1=a0a1+1a1 [an,a1,,aN]=[a0,a1,an1,[an,an+1,,aN]] =[an,an+1,an]pn1+pn2[an,an+1,an]qn1+qn2 x=anpn1+pn2anqn1+qn2\begin{gathered} x=a_{0}^{\prime}, \quad x=a_{0}+\frac{1}{a_{1}^{\prime}}=\frac{a_{0} a_{1}^{\prime}+1}{a_{1}^{\prime}} \\ \ \\ \left[a_{n}, a_{1}, \ldots, a_{N}\right]=\left[a_{0}, a_{1}, \ldots a_{n-1},\left[a_{n}, a_{n+1}, \dots, a_{N}\right]\right] \\ \ \\ =\frac{\left[a_{n}, a_{n+1} \ldots, a_{n}\right] p_{n-1}+p_{n-2}}{\left[a_{n}, a_{n+1} \dots, a_{n}\right] q_{n-1}+q_{n-2}} \\ \ \\ x=\frac{a_{n}' p_{n-1}+p_{n-2}}{a_{n}' q_{n-1}+q_{n-2}} \end{gathered}

theorem3\textbf{theorem3}
除了 aN=1,aN1=[aN1]1\text{除了} \ a_N = 1, \textbf{有} a_{N-1} = [a_{N-1}']-1
其他情况, an=[an]\text{其他情况, }a_n = [a_n']

注意到, an=an+1an+1(0nN1)\text{注意到, }a_{n}^{\prime}=a_{n}+\frac{1}{a_{n+1}'} \quad(0 \leqslant n \leqslant N-1)

i)N=0,a0=a0=[a0] ii)N>0, n=N1,an+1=1,aN=1,我们有 aN1=aN1+1aN=aN1+1>1 from  n+1 down to n,an+1>1(0nN1) an<an<an+1(0nN1) an=[an](0nN1)\begin{gathered} \textbf{i)}\quad N = 0, a_0 = a_0'=[a_0'] \\ \ \\ \textbf{ii)}\quad N > 0, \ n = N-1, a_{n+1}^{\prime}=1, a_{N}=1, \text{我们有} \\ \ \\ a_{N-1}^{\prime}=a_{N-1}+\frac{1}{a_{N}^{\prime}}=a_{N-1}+1>1 \\ \ \\ \text{from } \ n + 1 \ \text{down to} \ n, a_{n+1}^{\prime}>1 \quad(0 \leqslant n \leqslant N-1) \\ \ \\ a_{n}<a_{n}^{\prime}<a_{n}+1 \quad(0 \leqslant n \leqslant N-1) \\ \ \\ a_{n}=\left[a_{n}^{\prime}\right] \quad(0 \leqslant n \leq N-1) \end{gathered}

theorem4 连分数唯一展开\textbf{theorem4} \ \textbf{连分数唯一展开}
如果两个简单连分数
[a0,a1,,an1,an],[b0,b1,,bn1,bn]有同样的值 x[a_0, a_1, \ldots, a_{n-1}, a_n^{\prime}], [b_0, b_1, \ldots, b_{n-1}, b_n^{\prime}] \text{有同样的值 } x
aN>1,bM>1,M=N 并且这两个连分数完全相同a_N > 1, b_M > 1, M=N \ \text{并且这两个连分数完全相同}

i)n=m 根据 theorem3,a0=[x]=b0,那么  x=[a0,a1,,an1,an]=[a0,a1,,an1,bn] if n=1,a0+1a1=a0+1b1a1=b1a1=b1 anpn1+pn2anqn1+qn2=bnpn1+pn2bnqn1+qn2 (anbn)(pn1qn2pn2qn1)=0 pn1qn2pn2qn1=(1)n0an=bn an=bn\begin{gathered} \textbf{i)} \quad n = m \\ \ \\ \text{根据 }\text{theorem3}, a_0 = [x] = b_0, \text{那么 }\\ \ \\ x=\left[a_{0}, a_{1}, \ldots, a_{n-1}, a_{n}^{\prime}\right]=\left[a_{0}, a_{1}, \ldots, a_{n-1}, b_{n}^{\prime}\right] \\ \ \\ \text{if} \ n = 1, a_{0}+\frac{1}{a_{1}^{\prime}}=a_{0}+\frac{1}{b_{1}^{\prime}} \Rightarrow a_{1}^{\prime}=b_{1}^{\prime} \Rightarrow a_{1}=b_1 \\ \ \\ \frac{a_{n}^{\prime} p_{n-1}+p_{n-2}}{a_{n}^{\prime} q_{n-1}+q_{n-2}}=\frac{b_{n}^{\prime} p_{n-1}+p_{n-2}}{b_{n}^{\prime} q_{n-1}+q_{n-2}} \\ \ \\ \left(a_{n}^{\prime}-b_{n}^{\prime}\right)\left(p_{n-1} q_{n-2}-p_{n-2} q_{n-1}\right)=0 \\ \ \\ p_{n-1} q_{n-2}-p_{n-2} q_{n-1}=(-1)^{n} \neq 0 \Rightarrow a_{n}^{\prime}=b_{n}^{\prime} \\ \ \\ \Rightarrow a_{n}=b_{n} \end{gathered}

ii)NM,nNan=bn M>N,pNqN=x=[a0,a1,,aN]=[a0,a1,,aN,bN+1,,bM] =bN+1pN+pN1bN+1qN+qN1pNqN1pN1qN=0 这是不可能的  证明过程中, 我们假定了nN的部分, 所以 bn=[bN+1,,bM]\begin{gathered} \textbf{ii)} \quad N \leqslant M, \forall n \leqslant N \\ a_n = b_n \\ \ \\ M > N, \frac{p_{N}}{q_N}=x=\left[a_{0}, a_{1}, \ldots , a_{N}\right]=\left[a_{0}, a_{1}, \ldots, a_{N}, b_{N+1}, \ldots, b_{M}\right] \\ \ \\ =\frac{b_{N+1}^{\prime} p_{N}+p_{N-1}}{b_{N+1}^{\prime} q_{N}+ q_{N-1}} \Rightarrow p_{N}q_{ N-1}-p_{N-1} q_{N}=0 \\ \ \\ \text{这是不可能的 }\\ \ \\ \text{证明过程中, 我们假定了} n \leqslant N \text{的部分, 所以} \\ \ \\ b_{n}^{\prime}=\left[b_{N+1}, \ldots, b_{M}\right] \end{gathered}

Euclid算法

a0=[x],x=a0+ξ0(0ξ0<1)a_{0}=[x], \quad x=a_{0}+\xi_{0} \quad\left(0 \leqslant \xi_{0}<1\right)

1ξ0=a1,[a1]=a1,a1=a1+ξ1(0ξ1<1) 1ξ1=a2=a2+ξ2(0ξ2<1)\begin{gathered} \frac{1}{\xi_{0}}=a_{1}^{\prime}, \quad\left[a_{1}^{\prime}\right]=a_{1}, \quad a_{1}^{\prime}=a_{1}+\xi_{1} \quad\left(0 \leqslant \xi_{1}<1\right) \\ \ \\ \frac{1}{\xi_{1}}=a_{2}^{\prime}=a_{2}+\xi_{2} \quad\left(0 \leqslant \xi_{2}<1\right) \end{gathered}

a1>0,a2>0, a1>0,a2>0,x=a0+ξ0(0ξ0<1)1ξ0=a1=a1+ξ1(0ξ1<1)1ξ1=a2=a2+ξ2(0ξ2<1)\begin{gathered} a_{1}>0, \quad a_{2}>0, \quad \cdots \\ \ \\ \begin{gathered} &a_{1}>0, \quad a_{2}>0, \ldots\\ &x=a_{0}+\xi_{0} \quad\left(0 \leqslant \xi_{0}<1\right)\\ &\frac{1}{\xi_{0}}=a_{1}^{\prime}=a_{1}+\xi_{1} \quad\left(0 \leqslant \xi_{1}<1\right)\\ &\frac{1}{\xi_{1}}=a_{2}^{\prime}=a_{2}+\xi_{2} \quad\left(0 \leqslant \xi_{2}<1\right) \end{gathered} \end{gathered}