简介

József Wildt International Mathematical Competition（约瑟夫·维尔德国际数学竞赛）是一个面向高中生的国际数学竞赛，以纪念匈牙利数学家约瑟夫·维尔德（József Wildt）。它通常在罗马尼亚特尔古穆列什举办，旨在鼓励学生对数学的兴趣并挑战他们的解决问题的能力。

各届问题及其解答

2019 年

W1. The Pell numbers $P_n$ satisfy $P_0=0$, $P_1=1$, and $P_n=2P_{n-1}+P_{n-2}$ for $n\ge2$. Find

$\sum_{n=1}^{\infty}\left(\arctan\frac1{P_{2n}}+\arctan\frac1{P_{2n+2}}\right)\arctan\frac2{P_{2n+1}}.$

提示

依次证明:
(1). $P_{n}P_{n+2}+\left(-1\right)^{n}=P_{n+1}^{2}$;
(2). $\arctan\frac{1}{P_{2n}}-\arctan\frac{1}{P_{2n+2}}=\arctan\frac{2}{P_{2n+1}}$;
(3). $\sum\limits_{n=1}^{\infty}\left(\arctan\frac{1}{P_{2n}}+\arctan\frac{1}{P_{2n+2}}\right)\arctan\frac{2}{P_{2n+1}}=\arctan^{2}\frac{1}{2}$.

W2. If $0<a\le c\le b$ then:

$\frac{\left(b^{30}-a^{30}\right)\left(b^{30}-c^{30}\right)}{36b^{10}}\le\frac{\left(b^{25}-a^{25}\right)\left(b^{25}-c^{25}\right)}{25}\le\frac{\left(b^{30}-a^{30}\right)\left(b^{30}-c^{30}\right)}{36\left(ac\right)^{5}}$

提示

考虑

$\int_{a}^{b}\int_{c}^{b}t^{24}s^{24}\,ds\,dt\le\frac{1}{(ac)^{5}}\int_{a}^{b}\int_{c}^{b}t^{29}s^{29}\,ds\,dt$

和

$\frac{1}{b^{10}}\int_{a}^{b}\int_{c}^{b}t^{29}s^{29}\,ds\,dt\le\int_{a}^{b}\int_{c}^{b}t^{24}s^{24}\,ds\,dt.$

W3. Compute

$\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\frac{\cos x+1-x^{2}}{\left(1+x\sin x\right)\sqrt{1-x^{2}}}dx.$

提示

$\begin{align*} \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\frac{\cos x+1-x^{2}}{\left(1+x\sin x\right)\sqrt{1-x^{2}}}dx & =2\int_{0}^{\frac{\pi}{4}}\frac{\cos x+1-x^{2}}{\left(1+x\sin x\right)\sqrt{1-x^{2}}}dx\\ & =\left.-4\arctan\left(\frac{\cos\left(\frac{\arcsin x-x}{2}\right)}{\sin\left(\frac{\arcsin x+x}{2}\right)}\right)\right|_{0}^{\frac{\pi}{4}}\\ & =2\pi-4\arctan\left(\frac{\cos\left(\frac{\pi-4\arcsin\frac{\pi}{4}}{8}\right)}{\sin\left(\frac{\pi+4\arcsin\frac{\pi}{4}}{8}\right)}\right) \end{align*}$

W4. If $x,y,z,t>1$ then:

$\left(\log_{zxt}x\right)^{2}+\left(\log_{xyt}y\right)^{2}+\left(\log_{xyz}z\right)^{2}+\left(\log_{yzt}t\right)^{2}>\frac{1}{4}.$

提示

$\begin{align*} \sum_{cyc}\left(\frac{a}{a+c+d}\right)^{2} & >\frac{1}{4}\\ \sum_{cyc}1\sum_{cyc}\left(\frac{a}{a+c+d}\right)^{2} & \ge\left(\sum_{cyc}\frac{a}{a+c+d}\right)^{2}>\left(\sum_{cyc}\frac{a}{a+b+c+d}\right)^{2}=1. \end{align*}$

W5. Let $n\ge 1$. Find a set of distincts real numbers $(x_j)_{1\le j\le n}$ such that for any bijections $f:\{1,2,\cdots, n\}^2\to\{1,2,\cdots,n\}^2$ the matrix $\left(x_{f(i,j)}\right)_{1\le i,j\le n}$ is invertible.

W6. Compute

$\int_{\frac{\pi}{6}}^{\frac{\pi}{4}}\frac{(1+\ln x)\cos x+x\sin x\ln x}{\cos^{2}x+x^{2}\ln^{2}x}\ud x$

提示

$\begin{align*} \int_{\frac{\pi}{6}}^{\frac{\pi}{4}}\frac{(1+\ln x)\cos x+x\sin x\ln x}{\cos^{2}x+x^{2}\ln^{2}x}dx & =\int_{\frac{\pi}{6}}^{\frac{\pi}{4}}\frac{(x\ln x)^{\prime}\cos x+\sin x\cdot x\ln x}{\cos^{2}x+x^{2}\ln^{2}x}dx\\ & =\int_{\frac{\pi}{6}}^{\frac{\pi}{4}}\frac{1}{1+\left(\frac{x\ln x}{\cos x}\right)^{2}}d\left(\frac{x\ln x}{\cos x}\right)\\ & =\arctan\left(\frac{x\ln x}{\cos x}\right)|_{\pi/6}^{\pi/4}\\ & =\arctan\left(\frac{\sqrt{2}\pi\ln\frac{\pi}{4}}{4}\right)-\arctan\left(\frac{\sqrt{3}\pi\ln\frac{\pi}{6}}{9}\right) \end{align*}$

W7. If

$\Omega_n=\sum_{k=1}^n\left(\int_{-\frac{1}{k}}^{\frac{1}{k}}\left(2 x^{10}+3 x^8+1\right) \cdot \cos ^{-1}(k x) d x\right)$

then find:

$\Omega=\lim _{n \rightarrow \infty}\left(\Omega_n-\pi \cdot H_n\right)$

提示

分部积分得到:

$\int_{-\frac{1}{k}}^{\frac{1}{k}}x^{2n}\arccos\left(kx\right)\ud x=\frac{\pi}{(2n+1)k^{2n+1}},$

代入,

$\Omega_{n}=\sum_{k=1}^{n}\left(\int_{-\frac{1}{k}}^{\frac{1}{k}}\left(2x^{10}+3x^{8}+1\right)\cdot\arccos(kx)dx\right)=\sum_{k=1}^{n}\left(\frac{2\pi}{11k^{11}}+\frac{\pi}{3k^{9}}+\frac{\pi}{k}\right),$

所以

$\Omega=\frac{2\pi}{11}\zeta(11)+\frac{\pi}{3}\zeta(9).$

W8. Let $(a_n)_{n \geq 1}$ be a positive real sequence given by $a_n=\sum_{k=1}^n \frac{1}{k}$. Compute

$\lim _{n \to \infty} e^{-2 a_n} \sum_{k=1}^n\left[(\sqrt[2 k]{k!}+\sqrt[2(k+1)]{(k+1)!})^2\right]$

where we denote by $[x]$ the integer part of $x$.

提示

由 Stirling 公式,

$n!=\sqrt{2\pi n}\left(\frac{n}{\ue}\right)^{n}\ue^{O\left(\frac{1}{n}\right)},$

所以

$\sqrt[n]{n!}=\frac{n}{\ue}\cdot\ue^{\frac{\ln\left(2\pi n\right)}{2n}+O\left(\frac{1}{n^{2}}\right)}=\frac{n}{\ue}\left(1+\frac{\ln\left(2\pi n\right)}{2n}+o\left(\frac{1}{n}\right)\right),$

又由于不等式

$\sqrt[k]{k!}<\sqrt[k+1]{(k+1)!}\Longrightarrow\sqrt[2k]{k!}<\sqrt[2(k+1)]{(k+1)!},$

所以有自然的不等式 (下界有稍微的不严谨, 不过影响不大, 这里忽略)

$4\sqrt[k]{k!}<\left[\left(\sqrt[2k]{k!}+\sqrt[2(k+1)]{(k+1)!}\right)^{2}\right]<4\sqrt[k+1]{(k+1)!}.$

所以所求极限化为

$\lim_{n\to\infty}4\ue^{-2H_{n}}\sum_{k=1}^{n}\sqrt[k]{k!}<\lim_{n\to\infty}\ue^{-2H_{n}}\sum_{k=1}^{n}\left[\left(\sqrt[2k]{k!}+\sqrt[2(k+1)]{(k+1)!}\right)^{2}\right]<\lim_{n\to\infty}4\ue^{-2H_{n}}\sum_{k=1}^{n}\sqrt[k+1]{(k+1)!}.$

由于 $\ue^{-2H_{n}}=\ue^{-2(H_{n}-\ln n)-2\ln n}\to\ue^{-2\gamma}\cdot n^{-2}$,
所以

$\begin{align*} \lim_{n\to\infty}4\ue^{-2H_{n}}\sum_{k=1}^{n}\sqrt[k]{k!} & =\lim_{n\to\infty}\frac{4}{\ue^{2\gamma}}\cdot\frac{1}{n}\sum_{k=1}^{n}\frac{\sqrt[k]{k!}}{n}\\ & =\lim_{n\to\infty}\frac{4}{\ue^{2\gamma}}\cdot\frac{1}{n}\sum_{k=1}^{n}\frac{\frac{k}{\ue}\left(1+\frac{\ln\left(2\pi k\right)}{2k}+o\left(\frac{1}{k}\right)\right)}{n}\\ & =\frac{4}{\ue^{2\gamma}}\cdot\int_{0}^{1}\frac{x}{\ue}\ud x+\lim_{n\to\infty}\frac{4}{\ue^{2\gamma}}\cdot\frac{1}{n}\sum_{k=1}^{n}\left(\frac{\ln\left(2\pi k\right)}{2kn}+o\left(\frac{1}{kn}\right)\right)\\ & =\frac{2}{\ue^{2\gamma+1}}. \end{align*}$

同理可证:

$\lim_{n\to\infty}4\ue^{-2H_{n}}\sum_{k=1}^{n}\sqrt[k+1]{(k+1)!}=\frac{2}{\ue^{2\gamma+1}}.$

所以

$\lim_{n\to\infty}\ue^{-2H_{n}}\sum_{k=1}^{n}\left[\left(\sqrt[2k]{k!}+\sqrt[2(k+1)]{(k+1)!}\right)^{2}\right]=\frac{2}{\ue^{2\gamma+1}}.$

W9. Let $\alpha>0$ be a real number. Compute the limit of the sequence $\left\{x_n\right\}_{n \geq 1}$ defined by

$x_n=\left\{\begin{array}{ll} \sum_{k=1}^n \sinh \left(\frac{k}{n^2}\right), & n>\frac{1}{\alpha} ; \\ 0, & n \leq \frac{1}{\alpha} \end{array} .\right.$

提示

用 Taylor 展开 $\sinh x=x+\frac{x^{3}}{6}+o\left(u^{3}\right)$, 结果为 $1/2$.

W10. If $\mathrm{si}(x)=-\int_x^{\infty}\left(\frac{\sin t}{t}\right) \ud t$, $x>0$, then:

$\int_{\ue}^{\ue^{2}}\frac{\mathrm{si}(\ue^{4}x)-\mathrm{si}(\ue^{3}x)}{x}\ud x=\int_{\ue^{3}}^{\ue^{4}}\frac{\mathrm{si}(\ue^{2}x)-\mathrm{si}(\ue x)}{x}\ud x.$

提示

这本质上是一道交换积分次序问题, 注意到

$\mathrm{si}(x)=-\frac{\pi}{2}+\int_{0}^{x}\frac{\sin t}{t}\ud t,$

所以等式的左边为

$\begin{align*} \int_{\ue}^{\ue^{2}}\frac{\mathrm{si}(\ue^{4}x)-\mathrm{si}(\ue^{3}x)}{x}\ud x & =\int_{\ue}^{\ue^{2}}\int_{\ue^{3}x}^{\ue^{4}x}\frac{\sin t}{tx}\ud t\ud x\\ & =\int_{\ue^{4}}^{\ue^{6}}\left(\int_{\ue}^{t/\ue^{3}}\frac{\sin t}{tx}\ud x+\int_{t/\ue^{4}}^{\ue^{2}}\frac{\sin t}{tx}\ud x\right)\ud t\\ & =\int_{\ue^{4}}^{\ue^{6}}\frac{2\sin t}{t}\ud t; \end{align*}$

同理, 等式的右边为

$\begin{align*} \int_{\ue^{3}}^{\ue^{4}}\frac{\mathrm{si}(\ue^{2}x)-\mathrm{si}(\ue x)}{x}\ud x & =\int_{\ue^{3}}^{\ue^{4}}\int_{\ue x}^{\ue^{2}x}\frac{\sin t}{tx}\ud t\ud x\\ & =\int_{\ue^{4}}^{\ue^{6}}\left(\int_{\ue^{3}}^{t/\ue}\frac{\sin t}{tx}\ud x+\int_{t/\ue^{2}}^{\ue^{4}}\frac{\sin t}{tx}\ud x\right)\ud t\\ & =\int_{\ue^{4}}^{\ue^{6}}\frac{2\sin t}{t}\ud t. \end{align*}$

W11. Let $\left(s_n\right)_{n \geq 1}$ be a sequence given by $s_n=-2 \sqrt{n}+\sum_{k=1}^n \frac{1}{\sqrt{k}}$ with $\lim\limits_{n \to \infty} s_n=s=$ Ioachimescu constant and $\left(a_n\right)_{n \geq 1}$, $\left(b_n\right)_{n \geq 1}$ be a positive real sequences such that

$\lim_{n \to \infty} \frac{a_{n+1}}{n a_n}=a \in R_{+}^*, \quad\lim_{n \to \infty} \frac{b_{n+1}}{b_n \sqrt{n}}=b \in R_{+}^*$

Compute

$\lim_{n \to \infty}\left(1+e^{s_n}-e^{s_{n+1}}\right)^{\sqrt[n]{a_n b_n}}$

提示

由

$\lim_{n\to\infty}\frac{a_{n}}{na_{n-1}}=a,\quad\lim_{n\to\infty}\frac{b_{n}}{b_{n-1}\sqrt{n}}=b,$

所以有

$\lim_{n\to\infty}\frac{a_{n}b_{n}}{a_{n-1}b_{n-1}\cdot n^{3/2}}=ab,$

取对数, 做 Cesaro 平均的极限有

$\lim_{n\to\infty}\left(\frac{\ln(a_{n}b_{n})-\frac{3}{2}\ln(n!)}{n}\right)=\ln(ab),$

所以

$\ln\sqrt[n]{a_{n}b_{n}}=\frac{3}{2}\ln\left(\frac{n}{\ue}\right)+\ln(ab)+o(1)\Longrightarrow\sqrt[n]{a_{n}b_{n}}\sim\left(\frac{n}{\ue}\right)^{3/2}ab.$

由于 $\lim s_{n}=s$ 存在, 所以 $\lim\left(\ue^{s_{n}}-\ue^{s_{n+1}}\right)=0$, 有等价无穷小代换 $\ln\left(1+\ue^{s_{n}}-\ue^{s_{n+1}}\right)\sim\ue^{s_{n}}\left(1-\ue^{s_{n+1}-s_{n}}\right)$, 对后一项做幂级数展开可得:

$1-\ue^{s_{n+1}-s_{n}}=1-\ue^{-\frac{2}{\sqrt{n+1}+\sqrt{n}}+\frac{1}{\sqrt{n+1}}}=1-\ue^{-\frac{1}{4n\sqrt{n}}+o\left(\frac{1}{n\sqrt{n}}\right)}=\frac{1}{4n\sqrt{n}}+o\left(\frac{1}{n\sqrt{n}}\right).$

因 $\ue^{s_{n}}\to\ue^{s}$, 所以原极限等于

$\lim_{n\to\infty}\ue^{\ln\left(1+\ue^{s_{n}}-\ue^{s_{n+1}}\right)\cdot\sqrt[n]{a_{n}b_{n}}}=\lim_{n\to\infty}\ue^{\ue^{s}\left(\frac{1}{4n\sqrt{n}}+o\left(\frac{1}{n\sqrt{n}}\right)\right)\cdot ab\left(\frac{n}{\ue}\right)^{3/2}}=\ue^{\frac{ab\ue^{s-3/2}}{4}}.$

W12. If $0<a<b$ then:

$\frac{\int_a^{\frac{a+b}{2}}\left(\tan ^{-1} t\right) d t}{\int_a^b\left(\tan ^{-1} t\right) d t}<\frac{1}{2}$

提示

根据定积分的几何意义, 比较两个曲边梯形的面积.

W13. Let $a, b$ and $c$ be complex numbers such that $abc=1$. Find the value of the cubic root of

$\begin{vmatrix} b+n^3 c & n(c-b) & n^2(b-c) \\ n^2(c-a) & c+n^3 a & n(a-c) \\ n(b-a) & n^2(a-b) & a+n^3 b \end{vmatrix}.$

提示

结果为 $(n+1)(n^2-n+1)$.

W14. If $a, b, c>0$, $a b+b c+c a=3$ then:

$4\left(\tan ^{-1} 2\right)\left(\tan ^{-1}(\sqrt[3]{a b c})\right) \leq \pi \tan ^{-1}(1+\sqrt[3]{a b c}).$

提示

先证明: $1\ge abc\eqqcolon x$, 再用一阶导数与 $\left(1+x^{2}\right)\arctan x$ 的单调性, 证明函数 $\frac{\arctan(1+x)}{\arctan x}$ 单调递减.

W15. It is possible to partition the set $\{100,101, \cdots, 1000\}$ into two subsets so that for any two distinct elements $x$ and $y$ belonging to the same subset $\sqrt[3]{x+y}$ is irrational?

提示

这是一道线性代数问题. 考虑集合中任意两数和为立方数的所有可能结果为 $216, 343, 512, 729, 1000, 1331$, 现在需要划分两个集合, 使得同一集合中的任两个数之和不能是立方数. 此问题可以转化为和为立方数的两个整数应位于不同的集合, 求出这两个和为立方数的所有可能整数, 比如 $801+530=1331=11^3$, 所以 $801$ 与 $530$ 不在同一个集合内.

要确定集合划分为两个子集的方式是否存在, 这等价于验证是否有三个数 $x,y,z\in\{100,101, \cdots, 1000\}$, 使得它们任何两个数之和同时为立方数. 因为如果 $x,y,z$ 中的任何两个数之和为立方数, 则它们两两不能在一个集合内.

求解方程 $x+y=729$, $y+z=1000$, $z+x=1331$, 得到 $x=530$, $y=199$, $z=801$, 这三个数两两不能同时位于一个集合内, 因此集合的二划分不存在.

W16. If $f:[a, b] \to (0, \infty)$. $0<a \le b$. $f$ derivable, $f’$ continuous, then:

$\int_a^b \frac{f^{\prime}(x) \sqrt{f(x)}}{f^3(x)+1} d x \leq \tan ^{-1}\left(\frac{f(b)-f(a)}{1+f(a) f(b)}\right).$

提示

用凑微分法以及三角恒等式, 不等式等价于

$\frac{2}{3}\left(\arctan\sqrt{f^{3}(b)}-\arctan\sqrt{f^{3}(a)}\right)\le\arctan f(b)-\arctan f(a),$

然后用 Cauchy 中值定理证明

$\frac{\arctan\sqrt{f^{3}(b)}-\arctan\sqrt{f^{3}(a)}}{\arctan f(b)-\arctan f(a)}\le\frac{3}{2}.$

W17. Let $f_n=\left(1+\frac{1}{n}\right)^n\left((2 n-1)!!F_n\right)^{1 / n}$. Find $\lim\limits_{n \to \infty}\left(f_{n+1}-f_n\right)$, where $F_n$ denotes the $n$th Fibonacci number (given by $F_0=0$, $F_1=1$, and by $F_{n+1}=F_n+F_{n-1}$ for all $n \ge 1$).

提示

用幂级数展开, 给出每一个因子的较为精确的估计:

$\left(1+\frac{1}{n}\right)^{n}=\ue^{n\ln\left(1+\frac{1}{n}\right)}=\ue\left(1-\frac{1}{2n}+O\left(\frac{1}{n^{2}}\right)\right),$

和

$\begin{align*} \left((2n-1)!!F_{n}\right)^{1/n} & =\ue^{\frac{1}{n}\left(\ln(2n-1)!!+\ln F_{n}\right)}\\ & =\ue^{\ln(2\phi n)-1+\frac{\ln(2/5)}{2n}+O\left(\frac{1}{n^{2}}\right)}\\ & =\frac{2\phi n}{\ue}+\frac{\phi}{\ue}\ln\frac{2}{5}+O\left(\frac{1}{n}\right) \end{align*}$

所以

$\begin{align*} f_{n} & =\ue\left(1-\frac{1}{2n}+O\left(\frac{1}{n^{2}}\right)\right)\cdot\left(\frac{2\phi n}{\ue}+\frac{\phi}{\ue}\ln\frac{2}{5}+O\left(\frac{1}{n}\right)\right)\\ & =2\phi n+\phi\ln\frac{2}{5}-\phi+O\left(\frac{1}{n}\right), \end{align*}$

代入原极限, 便可得

$\lim_{n\to\infty}\left(f_{n+1}-f_{n}\right)=\lim_{n\to\infty}\left(2\phi+O\left(\frac{1}{n}\right)\right)=2\phi=1+\sqrt{5}.$

W18. Let $\left\{c_k\right\}_{k \ge 1}$ be a sequence with $0 \le c_k \le 1$, $c_1 \ne 0$, $\alpha>1$. Let $C_n=c_1+\cdots+c_n$. Prove

$\lim_{n \to \infty} \frac{C_1^\alpha+\cdots+C_n^\alpha}{\left(C_1+\cdots+C_n\right)^\alpha}=0.$

提示

当 $\left\{ C_{n}\right\} $ 收敛于某个常数 $C$, 使用常见的不等式 $\left(x^{\alpha}+y^{\alpha}\right)^{1/\alpha}<x+y$, $x^{\alpha}+y^{\alpha}<\left(x+y\right)^{\alpha}$.

取 $\epsilon=\frac{C}{2}$, $\exists N$, s.t. $\forall n>N$, $\left|C_{n}-C\right|<\epsilon$, 也即 $\frac{3C}{2}>C_{n}>\frac{C}{2}$. 再用夹逼原理:

$0<\frac{C_{1}^{\alpha}+\cdots+C_{N}^{\alpha}+\cdots+C_{n}^{\alpha}}{\left(C_{1}+\cdots+C_{N}+\cdots+C_{n}\right)^{\alpha}}<\frac{C(N)+(n-N)\left(\frac{3C}{2}\right)^{\alpha}}{\left(C(N)+(n-N)\frac{C}{2}\right)^{\alpha}}<\frac{C(N)+nC}{C(N)+n^{\alpha}C}\to0.$

若 $\left\{ C_{n}\right\} $ 发散于无穷大, 用 Stolz 公式

$\begin{align*} \lim_{n\to\infty}\frac{\left(C_{1}^{\alpha}+\cdots+C_{n}^{\alpha}\right)^{1/\alpha}}{C_{1}+\cdots+C_{n}} & =\lim_{n\to\infty}\frac{\left(1+\frac{C_{n}^{\alpha}}{C_{1}^{\alpha}+\cdots+C_{n-1}^{\alpha}}\right)^{1/\alpha}-1}{C_{n}}\left(C_{1}^{\alpha}+\cdots+C_{n-1}^{\alpha}\right)^{1/\alpha}\\ & =\lim_{n\to\infty}\left[\left(1+\frac{C_{n}^{\alpha}}{C_{1}^{\alpha}+\cdots+C_{n-1}^{\alpha}}\right)^{1/\alpha}-1\right]\left(\frac{C_{1}^{\alpha}+\cdots+C_{n-1}^{\alpha}}{C_{n}^{\alpha}}\right)^{1/\alpha} \end{align*}.$

再用一次 Stolz 公式, 得到:

$\lim_{n\to\infty}\left(\frac{C_{n}^{\alpha}}{C_{1}^{\alpha}+\cdots+C_{n-1}^{\alpha}}\right)=\lim_{n\to\infty}\frac{C_{n+1}^{\alpha}-C_{n}^{\alpha}}{C_{n}^{\alpha}}=\lim_{n\to\infty}\left(1+\frac{c_{n+1}}{C_{n}}\right)^{\alpha}-1=0,$

所以

$\left(1+\frac{C_{n}^{\alpha}}{C_{1}^{\alpha}+\cdots+C_{n-1}^{\alpha}}\right)^{1/\alpha}-1\sim\frac{C_{n}^{\alpha}}{\alpha\left(C_{1}^{\alpha}+\cdots+C_{n-1}^{\alpha}\right)}.$

原极限化为

$\begin{align*} \lim_{n\to\infty}\frac{\left(C_{1}^{\alpha}+\cdots+C_{n}^{\alpha}\right)^{1/\alpha}}{C_{1}+\cdots+C_{n}} & =\lim_{n\to\infty}\left[\left(1+\frac{C_{n}^{\alpha}}{C_{1}^{\alpha}+\cdots+C_{n-1}^{\alpha}}\right)^{1/\alpha}-1\right]\left(\frac{C_{1}^{\alpha}+\cdots+C_{n-1}^{\alpha}}{C_{n}^{\alpha}}\right)^{1/\alpha}\\ & =\lim_{n\to\infty}\left[\frac{C_{n}^{\alpha}}{\alpha\left(C_{1}^{\alpha}+\cdots+C_{n-1}^{\alpha}\right)}\right]\left(\frac{C_{1}^{\alpha}+\cdots+C_{n-1}^{\alpha}}{C_{n}^{\alpha}}\right)^{1/\alpha}\\ & =\lim_{n\to\infty}\left[\frac{C_{n}^{\alpha}}{\alpha\left(C_{1}^{\alpha}+\cdots+C_{n-1}^{\alpha}\right)}\right]^{1-\frac{1}{\alpha}}=\left(\frac{0}{\alpha}\right)^{1-\frac{1}{\alpha}}=0. \end{align*}$

W19. Let $\left\{F_n\right\}_{n \in \mathbb{Z}}$ and $\left\{L_n\right\}_{n \in \mathbb{Z}}$ denote the Fibonacci and Lucas numbers, respectively, given by

$F_{n+1}=F_n+F_{n-1} \text { and } L_{n+1}=L_n+L_{n-1} \text { for all } n \ge 1,$

with $F_0=0$, $F_1=1$, $L_0=2$, and $L_1=1$. Prove that for integers $n \ge 1$ and $j \ge 0$

(i) $\sum\limits_{k=1}^n F_{k \pm j} L_{k \mp j}=F_{2 n+1}-1+\left\{\begin{array}{ll}0, & \text { if } n \text { is even } \\ (-1)^{ \pm j} F_{ \pm 2 j}, & \text { if } n \text { is odd }\end{array}\right.$.

(ii) $\sum\limits_{k=1}^n F_{k+j} F_{k-j} L_{k+j} L_{k-j}=\frac{F_{4 n+2}-1-n L_{4 j}}{5}$.

提示

设特征方程 $x^{2}=x+1$ 的两个解分别为 $x_{1},x_{2}$, 满足韦达定理: $x_{1}+x_{2}=1$, $x_{1}x_{2}=-1$. 则有常数 $f_{1},f_{2},l_{1},l_{2}$ 使得$F_{n}=f_{1}x_{1}^{n}+f_{2}x_{2}^{n}$, $L_{n}=l_{1}x_{1}^{n}+l_{2}x_{2}^{n}$, $f_{1}+f_{2}=0$, $l_{1}+l_{2}=2$. 并且容易解得 $l_{1}=l_{2}=1$. 所以

$\begin{align*} \sum_{k=1}^{n}F_{k\pm j}L_{k\mp j} & =\sum_{k=1}^{n}\left(f_{1}x_{1}^{k\pm j}+f_{2}x_{2}^{k\pm j}\right)\left(x_{1}^{k\mp j}+x_{2}^{k\mp j}\right)\\ & =\sum_{k=1}^{n}\left(f_{1}x_{1}^{2k}+(-1)^{k\pm j}f_{1}x_{1}^{\pm2j}+(-1)^{k\pm j}f_{2}x_{2}^{\pm2j}+f_{2}x_{2}^{2k}\right)\\ & =-f_{1}x_{1}\left(1-x_{1}^{2n}\right)-f_{2}x_{2}\left(1-x_{2}^{2n}\right)+(-1)^{\pm j}\left(f_{1}x_{1}^{\pm2j}+f_{2}x_{2}^{\pm2j}\right)\sum_{k=1}^{n}(-1)^{k}\\ & =-f_{1}x_{1}-f_{2}x_{2}+f_{1}x_{1}^{2n+1}+f_{2}x_{2}^{2n+1}+(-1)^{j}F_{\pm2j}\sum_{k=1}^{n}\left(-1\right)^{k}\\ & =F_{2n+1}-F_{1}+(-1)^{j}F_{\pm2j}\sum_{k=1}^{n}\left(-1\right)^{k}. \end{align*}$

类似地, 有

$\begin{align*} \sum\limits_{k=1}^{n}F_{k+j}F_{k-j}L_{k+j}L_{k-j} & =\sum\limits_{k=1}^{n}\left(f_{1}x_{1}^{k+j}+f_{2}x_{2}^{k+j}\right)\left(f_{1}x_{1}^{k-j}+f_{2}x_{2}^{k-j}\right)\left(x_{1}^{k+j}+x_{2}^{k+j}\right)\left(x_{1}^{k-j}+x_{2}^{k-j}\right)\\ & =\sum\limits_{k=1}^{n}\left(f_{1}x_{1}^{2k}+f_{2}x_{2}^{2k}+f_{1}x_{1}^{k+j}x_{2}^{k-j}+f_{2}x_{2}^{k+j}x_{1}^{k-j}\right)\left(f_{1}x_{1}^{2k}+f_{2}x_{2}^{2k}+f_{1}x_{1}^{k-j}x_{2}^{k+j}+f_{2}x_{2}^{k-j}x_{1}^{k+j}\right)\\ & =\sum\limits_{k=1}^{n}\left(f_{1}x_{1}^{2k}+f_{2}x_{2}^{2k}+(-1)^{k-j}f_{1}x_{1}^{2j}+(-1)^{k-j}f_{2}x_{2}^{2j}\right)\left(f_{1}x_{1}^{2k}+f_{2}x_{2}^{2k}-(-1)^{k+j}f_{1}x_{1}^{2j}-(-1)^{k+j}f_{2}x_{2}^{2j}\right)\\ & =\sum\limits_{k=1}^{n}\left(\left(f_{1}x_{1}^{2k}+f_{2}x_{2}^{2k}\right)^{2}-\left((-1)^{k-j}f_{1}x_{1}^{2j}+(-1)^{k-j}f_{2}x_{2}^{2j}\right)^{2}\right)\\ & =\sum\limits_{k=1}^{n}\left(f_{1}^{2}x_{1}^{4k}+f_{2}^{2}x_{2}^{4k}+2f_{1}f_{2}-\left(f_{1}^{2}x_{1}^{4j}+f_{2}^{2}x_{2}^{4j}+2f_{1}f_{2}\right)\right)\\ & =\frac{f_{1}^{2}x_{1}^{4}(1-x_{1}^{4n})}{1-x_{1}^{4}}+\frac{f_{2}^{2}x_{2}^{4}(1-x_{2}^{4n})}{1-x_{2}^{4}}-n\left(f_{1}^{2}x_{1}^{4j}+f_{2}^{2}x_{2}^{4j}\right), \end{align*}$

然后用一点计算上的技巧计算 $\frac{f_{1}x_{1}^{2}}{1-x_{1}^{4}}=\frac{f_{2}x_{2}^{2}}{1-x_{2}^{4}}=-\frac{1}{5}$, $\frac{f_{1}^{2}x_{1}^{4}}{1-x_{1}^{4}}+\frac{f_{2}^{2}x_{2}^{4}}{1-x_{2}^{4}}=-\frac{1}{5}$, $f_{1}^{2}=f_{2}^{2}=\frac{1}{5}$, 其中 $f_{1}=-f_{2}=\frac{x_{1}}{x_{1}+2}$.

W20. i). Let $G$ be a $(4,4)$ unoriented graph, $2$-regulate, containing a cycle with the length $3$. Find the characteristic polynomial $P_G(\lambda)$, its spectrum $\mathrm{Spec}(G)$ and draw the graph $G$.

ii). Let $G^{\prime}$ be another $2$-regulate graph, having its characteristic polynomial $P_{G^{\prime}}(\lambda)=\lambda^4-4 \lambda^2+\alpha$, $\alpha \in R$. Find the spectrum $\mathrm{Spec}\left(G^{\prime}\right)$ and draw the graph $G^{\prime}$.

iii). Are the graphs $G$ and $G^{\prime}$ cospectral or isomorphic?

提示

$G$ 是一个有 4 个顶点 4 条边的无向图, 是 2 正则图 (即每个顶点的度均为 2), 图中有一个长度为 3 的圈. 则 G 的邻接矩阵为

$\mathrm{Adj}_G=\begin{pmatrix} 0 & 1 & 1 & 0\\ 1 & 0 & 1 & 0\\ 1 & 1 & 0 & 0\\ 0 & 0 & 0 & 1 \end{pmatrix}$

特征多项式为 $P_G(\lambda)=\lambda^4-\lambda^3-3\lambda^2+\lambda+2$, 谱为 $\mathrm{Spec}(G)=\{1,2,-1,-1\}$.

对于 2 正则图 $G^{\prime}$, 其顶点的度数均为 2, 且为特征多项式的一个特征值, 所以 $P_{G^{\prime}}(2)=0$, 解得 $\alpha=0$. 对于顶点数为 4 的 2 正则图, 除了自环 $C_1$ 和三个顶点的圈 $C_3$ 构成的 $C_1\cup C_3$ 外, 只有顶点数为 4 的圈 $C_4$, 而只有 $C_4$ 的特征多项式为 $\lambda^4-4\lambda^2$.

同构的图有相同特征多项式.

W21. Let $f$ be a continuously differentiable function on $[0,1]$ and $m \in \mathbb{N}$. Let $A=f(1)$ and let $B=\int_0^1 x^{-\frac{1}{m}} f(x) \ud x$. Calculate

$\lim_{n \to \infty} n\left(\int_0^1 f(x) \ud x-\sum_{k=1}^n\left(\frac{k^m}{n^m}-\frac{(k-1)^m}{n^m}\right) f\left(\frac{(k-1)^m}{n^m}\right)\right)$

in terms of $A$ and $B$.

提示

设 $x_{k}=\frac{k^{m}}{n^{m}}$, $x_{k-1}=\frac{(k-1)^{m}}{n^{m}}$, 则

$\sum_{k=1}^{n}\left(\frac{k^{m}}{n^{m}}-\frac{(k-1)^{m}}{n^{m}}\right)f\left(\frac{(k-1)^{m}}{n^{m}}\right)=\sum_{k=1}^{n}\Delta x_{k}f(x_{k-1}).$

所以

$\begin{align*} \int_{0}^{1}f(x)\ud x-\sum_{k=1}^{n}\left(\frac{k^{m}}{n^{m}}-\frac{(k-1)^{m}}{n^{m}}\right)f\left(\frac{(k-1)^{m}}{n^{m}}\right) & =\sum_{k=1}^{n}\int_{x_{k-1}}^{x_{k}}(f(x)-f(x_{k-1}))\ud x\\ & =\sum_{k=1}^{n}\left(\int_{x_{k-1}}^{x_{k}}f'(x_{k-1})(x-x_{k-1})+o(1)(x-x_{k-1})\ud x\right)\\ & =\sum_{k=1}^{n}\left(f'(x_{k-1})\cdot\frac{\Delta x_{k}^{2}}{2}+o(1)\frac{\Delta x_{k}^{2}}{2}\right), \end{align*}$

这里的无穷小量在 $n\to\infty$ 时一致收敛于零. 所以所求极限为

$\begin{align*} L=\lim_{n\to\infty}n\left(\cdots\right) & =\lim_{n\to\infty}\frac{1}{2}\sum_{k=1}^{n}\left(f'(x_{k-1})n\cdot\Delta x_{k}^{2}+o(1)n\cdot\Delta x_{k}^{2}\right)\\ & =\lim_{n\to\infty}\frac{1}{2}\sum_{k=1}^{n}\left(f'(x_{k-1})+o(1)\right)n\cdot\frac{m^{2}\xi_{k}^{2m-2}}{n^{2}}\\ & =\lim_{n\to\infty}\frac{m^{2}}{2n}\sum_{k=1}^{n}\left(f'(\xi_{k}^{m})+o(1)\right)\xi_{k}^{2m-2}\\ & =\frac{m^{2}}{2}\int_{0}^{1}x^{2m-2}f'(x^{m})\ud x\\ & =\frac{m^{2}}{2}\int_{0}^{1}\frac{t^{1-1/m}}{m}f'(t)\ud t\\ & =\frac{m}{2}A-\frac{m-1}{2}B, \end{align*}$

其中 $\xi_{k}\in\left(\frac{k-1}{n},\frac{k}{n}\right)$.

W22.: Let $A$ and $B$ the series:

$A=\sum_{n>0} \frac{C_{2 n}^1}{C_{2 n}^0+C_{2 n}^1+\ldots+C_{2 n}^{2 n}}, \quad B=\sum_{n>0} \frac{\Gamma\left(n+\frac{1}{2}\right)}{\Gamma\left(n+\frac{5}{2}\right)}.$

Study if $\frac{A}{B}$ is irrational number.

提示

$A=\frac89$, $B=\frac23$.

W23. If $b, c$ are the legs, and $a$ is the hypotenuse of a right triangle, prove that

$(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \geq 5+3 \sqrt{2}$

提示

用三角函数 $b=a\sin\alpha$, $c=a\cos\alpha$, 然后令 $t=\sin\alpha+\cos\alpha\in(1,\sqrt{2}]$, 有 $\sin\alpha\cos\alpha=\frac{t^2-1}{2}$.

W24. If $a, b, c>0$, prove that

$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \geq \frac{a+b}{a+b+2 c}+\frac{b+c}{b+c+2 a}+\frac{c+a}{c+a+2 b}$

提示

不等式等价于

$\sum_{cyc}\left(1+\frac{a}{b+c}\right)\ge\sum_{cyc}\left(1+\frac{a+b}{a+b+2c}\right)\Longleftrightarrow\sum_{cyc}\frac{1}{b+c}\ge\sum_{cyc}\frac{2}{a+b+2c}.$

这可以由 $\frac{1}{a}+\frac{1}{b}\ge\frac{2}{a+b}$ 证明.

W25. Let $x_i, y_i, z_i, \omega_i \in \RR^{+}, i=1,2 \cdots n$, such that

$\begin{aligned} & \sum_{i=1}^n x_i=n x, \sum_{i=1}^n y_i=n y, \sum_{i=1}^n \omega_i=n \omega \\ & \Gamma\left(z_i\right) \geqslant \Gamma\left(\omega_i\right), \sum_{i=1}^n \Gamma\left(z_i\right)=n \Gamma^*(z) \end{aligned}$

Then

$\sum_{i=1}^n \frac{\left(\Gamma\left(x_i\right)+\Gamma\left(y_i\right)\right)^2}{\Gamma\left(z_i\right)-\Gamma\left(\omega_i\right)} \geqslant n \frac{(\Gamma(x)+\Gamma(y))^2}{\Gamma^*(z)-\Gamma(\omega)} .$

提示

用 Titu 引理, (Cauchy-Schwarz 不等式的特殊情况), 即 $\forall a_{i},b_{i}>0$, 恒有

$\sum_{i}\frac{a_{i}^{2}}{b_{i}}\ge\frac{\left(\sum_{i}a_{i}\right)^{2}}{\sum_{i}b_{i}}.$

由于 $\Gamma$ 函数是下凸的, 由 Jensen 不等式, 有 $\frac{\sum\Gamma(x_{i})}{n}\ge\Gamma\left(\frac{\sum x_{i}}{n}\right)=\Gamma(x)$. 结合 Titu 引理, 有

$\begin{align*} \sum_{i}\frac{\left(\Gamma(x_{i})+\Gamma(y_{i})\right)^{2}}{\Gamma(z_{i})-\Gamma(w_{i})} & \ge\frac{\left(\sum_{i}(\Gamma(x_{i})+\Gamma(y_{i}))\right)^{2}}{\sum_{i}(\Gamma(z_{i})-\Gamma(w_{i}))}\\ & \ge\frac{\left(n(\Gamma(x)+\Gamma(y)\right)^{2}}{n(\Gamma^{*}(z)-\Gamma(w))}, \end{align*}$

化简即得.

W26. Let $n \in \NN$, $n \geq 2$, $a_1, a_2, \cdots, a_n \in \RR$ and $a_n=\max \left\{sa_1, a_2, \cdots, a_n\right\}$
a). If $t_k$, $t_k^{\prime} \in \RR$, $k \in\{1,2, \cdots, n\}$, $t_k \le t_k^{\prime}$, for any $k \in\{1,2, \cdots, n-1\}$ and

$\sum_{k=1}^n t_k=\sum_{k=1}^n t_k^{\prime},$

prove that

$\sum_{k=1}^n t_k a_k \geq \sum_{k=1}^n t_k^{\prime} a_k.$

b). If $b_k, c_k \in \RR_{+}^*, k \in\{1,2, \cdots, n\}, b_k \leq c_k$ for any $k \in\{1,2, \cdots, n-1\}$ and

$b_1 \cdot b_2 \cdot \cdots \cdot b_n=c_1 \cdot c_2 \cdot \cdots \cdot c_n,$

prove that

$\prod_{k=1}^n b_k^{a_k} \geq \prod_{k=1}^n c_k^{a_k}.$

提示

不等式等价于

$\sum_{k=1}^{n-1}\left(t_{k}'-t_{k}\right)\left(a_{n}-a_{k}\right)\ge0.$

W27. Find all continuous functions $f: \RR \rightarrow \RR$ such that

$f(-x)+\int_0^x t f(x-t) \ud t=x, \quad \forall x \in \RR.$

提示

$f(x)=-\sinh x$.

W28. In a room, we have 2019 aligned switches, connected to 2019 light bulbs, all initially switched on. Then, 2019 people enter the room one by one, performing the operation: The first, uses all the switches; the second, every second switch; the third, every third switch, and so on. How many lightbulbs remain switched on, after all the people entered?

提示

有 2019 个灯泡、2019 个开关排成一行。开始时所有灯泡都亮着（即开关处于“开”状态）。然后有 2019 个人依次进入，每个人都切换若干个开关：

第 1 个人切换所有开关；
第 2 个人切换每隔一个开关一次（编号为 2、4、6、…）；
第 3 个人切换每隔两个开关一次（编号为 3、6、9、…）；
…第 $k$ 人切换所有编号是 $k$ 的倍数的开关；
直到第 2019 个人完成操作。

问：最后有多少个灯泡仍然亮着？

灯泡 $n$ 被操作者切换的次数 $=n$ 的正约数个数 $d(n)$。

若灯泡被切换的次数是偶数($n$不是平方数)，则它最终仍是亮着。
若被切换的次数是奇数($n$是平方数)，则它最终是熄灭。

$亮着的灯泡数=2019−(小于等于 2019 的完全平方数个数)=2019-44=1975$.

W29. Prove that

$\int_0^{\infty} e^{3 t} \frac{4 e^{4 t}(3 t-1)+2 e^{2 t}(15 t-17)+18(1-t)}{\left(1+e^{4 t}-e^{2 t}\right)^2} d t=12 \sum_{k=0}^{\infty} \frac{(-1)^k}{(2 k+1)^2}-10.$

提示

用分部积分

$\begin{align*} \int_{0}^{\infty}\ue^{3t}\frac{4\ue^{4t}(3t-1)+2\ue^{2t}(15t-17)+18(1-t)}{(1+\ue^{4t}-\ue^{2t})^{2}}\ud t & =\frac{1}{2}\int_{0}^{\infty}\ue^{t}\frac{4\ue^{4t}(3t-1)+2\ue^{2t}(15t-17)+18(1-t)}{(1-2\ue^{2t})}\ud\frac{1}{\left(1+\ue^{4t}-\ue^{2t}\right)}\\ & =\frac{1}{2}\left[-20+18\int_{0}^{\infty}\frac{1}{1+\ue^{4t}-\ue^{2t}}\cdot t\ue^{t}\left(1+\ue^{2t}\right)\ud t\right]\\ & =-10+9\int_{0}^{\infty}\frac{t\ue^{-t}(1+\ue^{-2t})^{2}}{1+\ue^{-6t}}\ud t, \end{align*}$

用幂级数展开

$\begin{align*} \int_{0}^{\infty}\frac{t\ue^{-t}(1+\ue^{-2t})^{2}}{1+\ue^{-6t}}\ud t & =\sum_{k=0}^{\infty}(-1)^{k}\int_{0}^{\infty}t\ue^{-t}\left(1+\ue^{-2t}\right)^{2}\ue^{-6kt}\ud t\\ & =\sum_{k=0}^{\infty}\left(\frac{(-1)^{k}}{(6k+1)^{2}}+\frac{2(-1)^{k}}{(6k+3)^{2}}+\frac{(-1)^{k}}{(6k+5)^{2}}\right)\\ & =\frac{4}{3}\sum_{k=0}^{\infty}\frac{(-1)^{k}}{(2k+1)^{2}}. \end{align*}$

W30. a). Prove that

$\lim _{n \rightarrow \infty}\left(n+\frac{1}{4}-\zeta(3)-\zeta(5)-\cdots-\zeta(2 n+1)\right)=0,$

b). Calculate

$\sum_{n=1}^{\infty}\left(n+\frac{1}{4}-\zeta(3)-\zeta(5)-\cdots-\zeta(2 n+1)\right).$

提示

用交换求和次序,

$\begin{align*} n+\frac{1}{4}-\zeta(3)-\zeta(5)-\cdots-\zeta(2n+1) & =\sum_{k=2}^{\infty}\frac{1}{k^{2n}}\cdot\frac{1}{k(k^{2}-1)}\to0,\quad n\to\infty. \end{align*}$

所以

$\begin{align*} \sum_{n=1}^{\infty}\sum_{k=2}^{\infty}\frac{1}{k^{2n}}\cdot\frac{1}{k(k^{2}-1)} & =\sum_{k=2}^{\infty}\frac{1}{k(k^{2}-1)^{2}}\\ & =\sum_{k=2}^{\infty}\left(\frac{1}{k}-\frac{1}{2(k+1)}-\frac{1}{2(k-1)}\right)+\sum_{k=2}^{\infty}\left(\frac{1}{4(k-1)^{2}}-\frac{1}{4(k+1)^{2}}\right)\\ & =\frac{1}{16}. \end{align*}$

W31. Let $a, b \in \RR$, $a<b$ and the differentiable function $f:[a, b] \rightarrow \RR$, such that $f(a)=a$ and $f(b)=b$. Prove that

$\int_a^b\left(f^{\prime}(x)\right)^2 \ud x \ge b-a$

提示

用 Cauchy 不等式

$\int_{a}^{b}\ud x\int_{a}^{b}(f')^{2}\ud x\ge\left(\int_{a}^{b}\left|f'\right|\ud x\right)^{2}\ge\left(\int_{a}^{b}f'\ud x\right)^{2}=(b-a)^{2}.$

W32. Let $u_k$, $v_k$, $a_k$ and $b_k$ be non-negative real sequences such as $u_k>a_k$ and $v_k>b_k$, where $k=1,2, \cdots, n$. If $0<m_1 \le u_k \le M_1$ and $0<m_2 \le v_k \le M_2$, then

$\sum_{k=1}^n\left(\ell u_k v_k-a_k b_k\right) \ge\left(\sum_{k=1}^n\left(u_k^2-a_k^2\right)\right)^{1 / 2}\left(\sum_{k=1}^n\left(v_k^2-b_k^2\right)\right)^{1 / 2},$

where

$\ell=\frac{M_1 M_2+m_1 m_2}{2\left(m_1 M_1 m_2 M_2\right)^{1 / 2}}.$

提示

定理 (Pólya-Szegő 不等式): 设 $a=\left(a_{1},a_{2},\cdots,a_{n}\right)$, $b=\left(b_{1},b_{2},\cdots,b_{n}\right)$ 为两组正实数序列, 且 $0<m_{1}\le a_{k}\le M_{1}$, $0<m_{2}\le b_{k}\le M_{2}$. 则有

$\frac{\sum_{k=1}^{n}a_{k}^{2}\sum_{k=1}^{n}b_{k}^{2}}{\left(\sum_{k=1}^{n}a_{k}b_{k}\right)^{2}}\le\left(\frac{M_{1}M_{2}+m_{1}m_{2}}{2\sqrt{M_{1}M_{2}m_{1}m_{2}}}\right)^{2}.$

提示

证明: 设 $x_{k}=a_{k}/b_{k}$, 则有 $m\coloneqq\frac{m_{1}}{M_{2}}\le x_{k}\le\frac{M_{1}}{m_{2}}\eqqcolon M$. 因此

$\left(\frac{a_{k}}{b_{k}}-m\right)\left(\frac{a_{k}}{b_{k}}-M\right)\le0\Longleftrightarrow a_{k}^{2}+mMb_{k}^{2}\le(m+M)a_{k}b_{k},$

所以

$(m+M)\sum a_{k}b_{k}\ge\sum a_{k}^{2}+mM\sum b_{k}^{2}\ge2\sqrt{mM\sum a_{k}^{2}\sum b_{k}^{2}},$

也即

$\frac{\sum a_{k}^{2}\sum b_{k}^{2}}{\left(\sum a_{k}b_{k}\right)^{2}}\le\frac{(m+M)^{2}}{4mM}=\frac{(m_{1}m_{2}+M_{1}M_{2})^{2}}{4m_{1}m_{2}M_{1}M_{2}}.$

根据条件,

$\ell\sum u_{k}v_{k}\ge\sqrt{\sum u_{k}^{2}\sum v_{k}^{2}},$

再由 Cauchy 不等式, 有

$\sum(\ell u_{k}v_{k}-a_{k}b_{k})\ge\sqrt{\sum u_{k}^{2}\sum v_{k}^{2}}-\sqrt{\sum a_{k}^{2}\sum b_{k}^{2}}\ge\sqrt{\left(\sum u_{k}^{2}-\sum a_{k}^{2}\right)\left(\sum v_{k}^{2}-\sum b_{k}^{2}\right)}.$

注: 这个不等式以阿采尔不等式 (AcZél’s inequality)¹ 为推论: 如果 $x_{1}^{2}-\sum_{i=2}^{n}x_{i}^{2}>0$, $y_{1}^{2}-\sum_{i=2}^{n}y_{i}^{2}>0$, 则

$\left(x_{1}y_{1}-\sum_{i=2}^{n}x_{i}y_{i}\right)^{2}\ge\left(x_{1}^{2}-\sum_{i=2}^{n}x_{i}^{2}\right)\left(y_{1}^{2}-\sum_{i=2}^{n}y_{i}^{2}\right).$

证明只需取 $u_{k}\equiv\sqrt{\frac{2}{n}}x_{1}$, $v_{k}=\sqrt{\frac{2}{n}}y_{1}$, $a_{k}=x_{k}$, $b_{k}=y_{k}$.

注: Pólya-Szegő 不等式是卡萨马塔不等式（Cassels’ inequality）的特殊情况: 设 $a=\left(a_{1},\cdots,a_{n}\right)$, $b=\left(b_{1},\cdots,b_{n}\right)$ 是实数序列, $w=\left(w_{1},\cdots,w_{n}\right)$ 是正权重序列. 若

$0<m\le\frac{a_{k}}{b_{k}}\le M<\infty,\qquad k=1,2,\cdots,n,$

则有

$\frac{\left(\sum w_{k}a_{k}^{2}\right)\left(\sum w_{k}b_{k}^{2}\right)}{\left(\sum w_{k}a_{k}b_{k}\right)^{2}}\le\frac{(M+m)^{2}}{4Mm}.$

证明从 $\left(\frac{a_{k}}{b_{k}}-m\right)\left(\frac{a_{k}}{b_{k}}-M\right)\le0$ 得到 $w_{k}a_{k}^{2}+mMw_{k}b_{k}^{2}\le(M+m)w_{k}a_{k}b_{k}$, 所以

$(M+m)\sum w_{k}a_{k}b_{k}\ge\sum w_{k}a_{k}^{2}+mM\sum w_{k}b_{k}^{2}\ge2\sqrt{mM\sum w_{k}a_{k}^{2}\sum w_{k}b_{k}^{2}}.$

注: 卡萨马塔不等式（Cassels’ inequality）是康托洛维奇不等式（Kantorovich inequality）的推论: 若 $a_i>0$, $(i=1,2, \cdots, n)$ 且 $\sum_{i=1}^n a_i=1$，又 $0<\lambda_1 \le \lambda_2 \le \cdots \le \lambda_n$, 则

$\left(\sum_{i=1}^n \lambda_i a_i\right)\left(\sum_{i=1}^n \frac{a_i}{\lambda_i}\right) \le \frac{\left(\lambda_1+\lambda_n\right)^2}{4 \lambda_1 \lambda_n}.$

W33.² Let $0<\frac{1}{q} \le \frac{1}{p}<1$, and

$\frac1p+\frac1q=1.$

Let $u_k, v_k, a_k$ and $b_k$ be non-negative real sequences such as $u_k^p>a_k^p$ and $v_k^p>b_k^p$, where $k=1,2, \cdots, n$. If $0<m_1 \le u_k \le M_1$ and $0<m_2 \le v_k \le M_2$, then

$\left(\sum_{k=1}^{n}\left(\ell^{p}\left(u_{k}+v_{k}\right)^{p}-\left(a_{k}+b_{k}\right)^{p}\right)\right)^{1/p}\geq\left(\sum_{k=1}^{n}\left(u_{k}^{p}-a_{k}^{p}\right)\right)^{1/p}+\left(\sum_{k=1}^{n}\left(v_{k}^{p}-b_{k}^{p}\right)\right)^{1/p},$

where

$\ell=\frac{M_1 M_2+m_1 m_2}{2\left(m_1 M_1 m_2 M_2\right)^{1 / 2}}.$

W34. Let $a, b, c$ be positive real numbers and let $m, n$, ($m \ge n$) be positive integers. Prove that

$\frac{a^{n-1} b^{n-1} c^{m-n-1}}{a^{m+n}+b^{m+n}+a^n b^n c^{m-n}}+\frac{b^{n-1} c^{n-1} a^{m-n-1}}{b^{m+n}+c^{m+n}+b^n c^n a^{m-n}} +\frac{c^{n-1} a^{n-1} b^{m-n-1}}{c^{m+n}+a^{m+n}+c^n a^n b^{m-n}} \leq \frac{1}{a b c}$

提示

注意到

$\frac{a^{m}}{b^{n}}+\frac{b^{m}}{a^{n}}\ge\frac{a^{m}}{a^{n}}+\frac{b^{m}}{b^{n}},$

当且仅当 $a=b$ 时取等号. 所以

$\sum_{cyc}\frac{\frac{c^{m}}{c^{n}}}{\frac{c^{m}}{c^{n}}+\frac{a^{m}}{b^{n}}+\frac{b^{m}}{a^{n}}}\le\sum_{cyc}\frac{\frac{c^{m}}{c^{n}}}{\frac{c^{m}}{c^{n}}+\frac{a^{m}}{a^{n}}+\frac{b^{m}}{b^{n}}}=1.$

也即

$\sum_{cyc}\frac{a^{n-1}b^{n-1}c^{m-n-1}}{a^{m+n}+b^{m+n}+a^{n}b^{n}c^{m-n}}\le\frac{1}{abc},$

等号当且仅当 $a=b=c$ 时取到.

W35. Calculate

$\lim _{n \rightarrow \infty} \frac{n!\left(1+\frac{1}{n}\right)^{n^2+n}}{n^{n+1 / 2}}$

提示

用 Stirling 公式与 Taylor 展开, 答案是 $\sqrt{2\pi\ue}$.

W36. For any $a, b, c>0$ and for any $n \in \NN^*$, prove the inequality

$(a-b)\left(\frac{a}{b}\right)^n+(b-c)\left(\frac{b}{c}\right)^n+(c-a)\left(\frac{c}{a}\right)^n \geq(a-b) \frac{a}{b}+(b-c) \frac{b}{c}+(c-a) \frac{c}{a}.$

提示

不妨设 $a\ge b\ge c$, 则

$\mathop{\color{gray}\sum_{cyc}}\left(a-b\right)\left(\left(\frac{a}{b}\right)^{n}-\frac{a}{b}\right)\ge0.$

W37. For real $a>1$ find

$\lim _{n \rightarrow \infty} \sqrt[n]{\prod_{k=2}^n\left(a-a^{1 / k}\right)} .$

提示

取对数, 用 Stolz 公式, 答案是 $a-1$.