Mathematical Reflections 2023年第五期征解题

\section*{Mathematical Reflections 2023年第五期征解题}\section*{Junior Problems}\begin{flushright} 程国根\quad 译\end{flushright}{\bf J637.}求所有正整数$n$，使得$$3\cdot n!-1=\sqrt{(2n-1)!+1}.$$\vfill{\bf J638.}设$a,b,c$为正实数．求证：$$\frac{a+2 b}{\sqrt{c(a+2 b+3 c)}}+\frac{b+2 c}{\sqrt{a(b+2 c+3 a)}}+\frac{c+2 a}{\sqrt{b(c+2 a+3 b)}} \geq \frac{3 \sqrt{6}}{2}.$$\vfill{\bf J639.}解方程：$$176 x-4\lfloor x\rfloor^2-88\{x\}^2=2023,$$其中$\lfloor x\rfloor$和$\{x\}$分别表示$x$的整数部分和小数部分．\vfill{\bf J640.}正实数$x,y,z$满足$x+y+z=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$．求证：$$\sqrt{x y+3}+\sqrt{y z+3}+\sqrt{z x+3} \leq \frac{3(x+y+z+1)}{2}.$$\vfill{\bf J641.}求所有正整数$a,b$，使得$$\frac{a^2}{b}-\frac{b^2}{a}=2023.$$\vfill{\bf J642.}正实数$a,b,c$满足$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=3$．求证：$$3 \sqrt[6]{8 a b c} \leq \sqrt{a+\frac{b}{c}}+\sqrt{b+\frac{c}{a}}+\sqrt{c+\frac{a}{b}} \leq 3 \sqrt{2 a b c}.$$\vfill\newpage\section*{Mathematical Reflections 2023年第五期征解题}\section*{Senior Problems}\begin{flushright} 程国根\quad 译\end{flushright}{\bf S637.}设$a,b>0$，$c<0$，且满足$$2 a^2+\frac{1}{8 a^2}=7,2 b^2+\frac{1}{8 b^2}=17,2 c^2+\frac{1}{8 c^2}=31.$$求证：$$(4 a b+1)(4 b c+1)(4 c a+1)=(8 a b c+1)^2.$$\vfill{\bf S638.}设$a,b,c$为正实数．求证：$$\frac{6 a^2}{(2 b+c)(2 c+b)}+\frac{6 b^2}{(2 c+a)(2 a+c)}+\frac{6 c^2}{(2 a+b)(2 b+a)} \geq 1+\frac{a^2+b^2+c^2}{a b+b c+c a}.$$\vfill{\bf S639.}设$f(n)=\binom{n^{2}-1}{3}$，$n=2,3,4,\dots$求所有正整数$k\ge 2$，使得$f(k+3)=f(k)+2023$．\vfill{\bf S640.}设凸四边形$ABCD$有内切圆，圆心为$I$，半径为$r$．过$I$的一条直线分别和线段$BC,CA,AD$交于点$K,L,M$．设$r_{1}$是与线段$AB,AM,MK$均相切的圆的半径，$r_{2}$是与线段$CD,CK,KM$均相切的圆的半径．记$a=\dfrac{AL}{LC}$，求证：$$a\left(\frac{1}{r_{1}}-\frac{1}{r}\right)=\frac{1}{r_{2}}-\frac{1}{r}.$$\vfill{\bf S641.}求方程的整数解：$$2x^{3}-3x^{2}y^{2}+2y^{3}=1.$$\vfill{\bf S642.}正实数$a,b,c$满足$a+b+c=3$．求证：$$\frac{1}{a^2+3}+\frac{1}{b^2+3}+\frac{1}{c^2+3} \leq \frac{1}{92}\left(68+\frac{1}{a b c}\right) .$$\vfill\newpage\section*{Mathematical Reflections 2023年第五期征解题}\section*{Undergraduate Problems}\begin{flushright} 程国根\quad 译\end{flushright}{\bf U637.}设$H_{n}=\sum\limits_{k=1}^{n}\frac{1}{k}$．计算：$$\lim_{n\to\infty}n\left(\frac{n-H_{n}}{n}\right)^{n}.$$\vfill{\bf U638.}计算：$$\int\frac{1-\sin x}{3\sin x+5(1+\cos x)e^{x}}\mathrm{d}x$$其中$0<x<\dfrac{\pi}{2}$．\vfill{\bf U639.}设$a,b,c$为正实数，$a=\min\{a,b,c\}$且$a^{4}bc\ge 1$，令$$F(a,b,c)=\frac{a+b+c}{3}-\sqrt{\frac{ab+bc+ca}{3}}.$$求证：$$F(a,b,c)\ge F\left(\dfrac{1}{a},\dfrac{1}{b},\dfrac{1}{c}\right).$$\vfill{\bf U640.}数列$\{x_{n}\}_{n\ge 2}$定义如下：$x_{n}=\dfrac{\sqrt[n]{e}-1}{\sqrt[n^{2}]{e}-1}-n$．求证：$\lim\limits_{n\to\infty}x_{n}=\dfrac{1}{2}$．\vfill{\bf U641.}设$F(x,y,z)$为有理系数多项式．求证：存在有理系数多项式$Q(x,y,z)$和$R(x,y,z)$，使得$$P(x,y,z)Q(x,y,z)=R(x^{2}y,y^{2}z,z^{2}x).$$\vfill{\bf U642.}计算：$$\lim_{n\to\infty}n\sin ((2\pi n)^{p}+2^{p}\pi^{p}n^{p-1})^{\frac{1}{p}}.$$\vfill\newpage\section*{Mathematical Reflections 2023年第五期征解题}\section*{Olympiad Problems}\begin{flushright} 程国根\quad 译\end{flushright}{\bf O637.}正实数$a,b,c$满足$a+b+c=ab+bc+ca$．求证：$$\sqrt[3]{a^{3}+7}+\sqrt[3]{b^{3}+7}+\sqrt[3]{c^{3}+7}\le 2(a+b+c).$$\vfill{\bf O638.}设$1\le a_{1}<a_{2}<\dots$为无穷正整数列．求证：存在正整数列$b_{1},b_{2},\dots$，满足$b_{i}>a_{i}$，且满足$f(b_{i}+b_{j})=f(b_{i})+f(b_{j})$的唯一积性函数$f:\mathbb{N}\to\mathbb{Z}-\{0\}$为$f(n)=n$．\vfill{\bf O639.}正实数$a,b,c,\lambda$满足$$\frac{1}{a+\lambda}+\frac{1}{b+\lambda}+\frac{1}{c+\lambda}\le\frac{1}{\lambda}.$$求证：$$abc\ge 8\lambda^{3}\quad\text{且}\quad a+b+c+\frac{3abc}{ab+bc+ca}\ge 8\lambda.$$\vfill{\bf O640.}数列$\{a_{n}\}_{n\ge 0}$满足$a_{0}>1$，且对任意$n\ge 0$，都有$a_{n+1}=\dfrac{1+a_{n}^{2}}{a}$．求证：$$\prod_{k=0}^{n}\frac{1+a_{k}}{a_{k}}\ge \left(\frac{(1+a_{0})(n+1)}{(1+a_{0})n+a_{0}}\right)^{n+1}.$$\vfill{\bf O641.}设数列$\{F_{n}\}_{n\ge 0}$满足$F_{0}=0$，$F_{1}=1$，且对任意$n\ge 0$，都有$F_{n+2}=F_{n+1}+F_{n}$．$p,q$为两个不同的$4k+3$型素数．求所有三元非负整数组$(n,a,b)$，使得$F_{n}=p^{a}q^{b}$．\vfill{\bf O642.}点$P$在$\triangle ABC$内部．设$D=AP\cap BC$，$E=BP\cap CA$，$F=CP\cap AB$．$\triangle DEF$的外接圆和$BC,CA,AB$的另一个交点分别为$D',E',F'$．设$X=DF'\cap D'F$，$Y=EF'\cap E'F$．求证：$X,P,Y$三点共线．\vfill