播放地址
剧情简介
星期四��上21点，将会播出一集女����的����。这个����将会������都市��拉利�����������和����。女��们��供��在����上，其中也包����称为“美之女��”的����。��尔·������从���������生还后，本应该享��这场��会。然而，直到他收到一��来自��位����������的信，他��意��到这次女����可能会有些不同��常。信中提到，��尔��和他���自约会。在����都市的��个角��，少女��注一��地��出��定，这使得��尔和整个����都市都��入了������之中。与此同时，号称“最��”的“���������们”也开始行动了。这将是少年的���步，少女的期��-【��族之物语】。<|endoftext|> \begin{tikzpicture} %Definenodes \node[obs](x){$\mathbf{x}$}; \node[latent,above=ofx,xshift=-0.5cm](z){$\mathbf{z}$}; \node[latent,above=ofx,xshift=0.5cm](theta){$\boldsymbol{\theta}$}; \node[latent,below=ofx,xshift=0.75cm](w){$\mathbf{w}$}; %Connectthenodes \edge{z}{x} \edge{theta}{x} \edge{w}{x} %Plates \plate{plate1}{(x)(z)(theta)}{$N$}; \plate{plate2}{(x)(w)}{$M$}; \end{tikzpicture}<|endoftext|>Varjetalsomärsummanavtvåperfektakvadraterärdelbarmedfyra.<|endoftext|>##Problem1Evaluatethefollowingnumericalexpressions.a)$3+4$b)$7-2$c)$5-8$d)$12+10$a)$3+4=7$b)$7-2=5$c)$5-8=-3$d)$12+10=22$<|endoftext|>Algebraic<|endoftext|>#defineID1#defineNAME"JohnDoe"#defineAGE30#definePI3.14159265359<|endoftext|>##Problem1Let$a,b,c$bepositiverealnumberssuchthat$a+b+c=1$.Provethat$$\frac{a}{a^2+1}+\frac{b}{b^2+1}+\frac{c}{c^2+1}\leq\frac{9}{10}.$$Solution:Webeginbynotingthatforanypositiverealnumber$x$,wehave$$x^2+1\geq2x,$$sincethisinequalityisequivalentto$$x^2-2x+1=(x-1)^2\geq0,$$whichisclearlytrue.Hence,foranypositiverealnumbers$a,b,c$,wehave$$\frac{a}{a^2+1}\leq\frac{a}{2a}=\frac{1}{2}.$$Similarly,wehave$\frac{b}{b^2+1}\leq\frac{1}{2}$and$\frac{c}{c^2+1}\leq\frac{1}{2}$.Summingthesethreeinequalities,weobtain$$\frac{a}{a^2+1}+\frac{b}{b^2+1}+\frac{c}{c^2+1}\leq\frac{1}{2}+\frac{1}{2}+\frac{1}{2}=\frac{3}{2}.$$However,since$a+b+c=1$,wehave$a,b,c<1$,andhence$a^2,b^2,c^2<1$.Thisimpliesthat$a^2+1,b^2+1,c^2+1<2$,andtherefore,$\frac{1}{a^2+1},\frac{1}{b^2+1},\frac{1}{c^2+1}>\frac{1}{2}$.Multiplyingthesethreeinequalities,weobtain$$\frac{1}{(a^2+1)(b^2+1)(c^2+1)}>\frac{1}{2^3}=\frac{1}{8}.$$Raisingbothsidesofthisinequalitytothe$3$rdpower,weobtain$$\frac{1}{(a^2+1)^3(b^2+1)^3(c^2+1)^3}>\frac{1}{8^3}=\frac{1}{512}.$$Multiplyingbothsidesofthisinequalityby$a^3b^3c^3$,weobtain$$\frac{a^3b^3c^3}{(a^2+1)^3(b^2+1)^3(c^2+1)^3}>\frac{a^3b^3c^3}{512}.$$However,bytheAM-GMinequality,wehave$$\frac{a^3b^3c^3}{512}\leq\left(\frac{a+b+c}{3}\right)^9=\frac{1}{3^9}=\frac{1}{19683}.$$Combiningthisinequalitywiththepreviousone,weobtain$$\frac{a^3b^3c^3}{(a^2+1)^3(b^2+1)^3(c^2+1)^3}>\frac{1}{19683}.$$Finally,takingreciprocalsonbothsidesofthisinequality,weobtain$$(a^2+1)^3(b^2+1)^3(c^2+1)^3>19683,$$whichimpliesthat$$(a^2+1)(b^2+1)(c^2+1)>27.$$Hence,wehave$$\frac{1}{a^2+1}\cdot\frac{1}{b^2+1}\cdot\frac{1}{c^2+1}<\frac{1}{27}.$$Multiplyingbothsidesofthisinequalityby$a^2b^2c^2$,weobtain$$\frac{a^2b^2c^2}{(a^2+1)(b^2+1)(c^2+1)}<\frac{a^2b^2c^2}{27}.$$However,bytheAM-GMinequality,wehave$$\frac{a^2b^2c^2}{27}\leq\left(\frac{a+b+c}{3}\right)^6=\frac{1}{3^6}=\frac{1}{729}.$$Combiningthisinequalitywiththepreviousone,weobtain$$\frac{a^2b^2c^2}{(a^2+1)(b^2+1)(c^2+1)}<\frac{1}{729},$$andhence,$$\frac{a^2}{a^2+1}\cdot\frac{b^2}{b^2+1}\cdot\frac{c^2}{c^2+1}<\frac{1}{729}.$$Multiplyingbothsidesofthisinequalityby$a^2+b^2+c^2$,weobtain$$\frac{a^2}{a^2+1}\cdot\frac{b^2}{b^2+1}\cdot\frac{c^2}{c^2+1}\cdot(a^2+b^2+c^2)<\frac{a^2+b^2+c^2}{729}.$$However,bytheCauchy-Schwarzinequality,wehave$$(a^2+b^2+c^2)\left(\frac{a^2}{a^2+1}+\frac{b^2}{b^2+1}+\frac{c^2}{c^2+1}\right)\geq(a+b+c)^2=1,$$andhence,$$\frac{a^2}{a^2+1}+\frac{b^2}{b^2+1}+\frac{c^2}{c^2+1}\geq\frac{1}{a^2+b^2+c^2}>1.$$Combiningthisinequalitywiththepreviousone,weobtain$$\frac{a^2}{a^2+1}\cdot\frac{b^2}{b^2+1}\cdot\frac{c^2}{c^2+1}\cdot(a^2+b^2+c^2)<\frac{1}{729}\cdot\frac{a^2+b^2+c^2}{a^2+b^2+c^2}=\frac{1}{729}.$$Finally,multiplyingbothsidesofthisinequalityby$9$,weobtain$$\frac{9a^2}{a^2+1}\cdot\frac{9b^2}{b^2+1}\cdot\frac{9c^2}{c^2+1}<\frac{9}{729},$$whichsimplifiesto$$\frac{a^2}{a^2+1}\cdot\frac{b^2}{b^2+1}\cdot\frac{c^2}{c^2+1}<\frac{1}{81}.$$Combiningthisinequalitywiththeinequalityweobtainedearlier,weobtain$$\frac{a^2}{a^2+1}\cdot\frac{b^2}{b^2+1}\cdot\frac{c^2}{c^2+1}<\frac{1}{81}<\frac{1}{27}<\frac{1}{8}<\frac{1}{2}<\frac{3}{2}.$$Therefore,wehave$$\frac{a^2}{a^2+1}+\frac{b^2}{b^2+1}+\frac{c^2}{c^2+1}<\frac{3}{2}.$$Combiningthisinequalitywiththeinequalitiesweobtainedatthebeginningofthesolution,weobtain$$\frac{a}{a^2+1}+\frac{b}{b^2+1}+\frac{c}{c^2+1}<\frac{1}{2}+\frac{1}{2}+\frac{1}{2}=\frac{3}{2}<\frac{9}{10}.$$Hence,wehave$$\frac{a}{a^2+1}+\frac{b}{b^2+1}+\frac{c}{c^2+1}<\frac{9}{10},$$asdesired.$\square$<|endoftext|>#Language:Python3Notebook#Language:Python#1.Define
猜你喜欢
动漫周榜单
