基本不等式專題---完整版(非常全面)BasicInequalityTutorialI.SummaryofKnowledgePoints1.Originalformofbasicinequality(1)Ifa,b∈R,thena^2+b^2≥2ab(2)Ifa,b∈R,thenab≤(a^2+b^2)/22.Generalformofbasicinequality(meaninequality)Ifa,b∈R*,thena+b≥2√(ab)3.Twoimportanttransformationsofbasicinequality(1)Ifa,b∈R*,then(a+b)^2/4≥ab(2)Ifa,b∈R*,thenab≤[(a+b)/2]^2Summary:Whentheproductoftwopositivenumbersisconstant,theirsumhasaminimumvalue;whenthesumoftwopositivenumbersisconstant,theirproducthasaminimumvalue.Specialnote:Intheaboveinequalities,"="isonlyachievedwhena=b.4.Conditionsforfindingextremevalues:"Onepositive,twofixed,threeequal"5.Commonconclusions(1)Ifx>0,thenx+1/x≥2(achievedonlywhenx=1)(2)Ifx<0,thenx+1/x≤-2(achievedonlywhenx=-1)(3)Ifab>0,thena/b+b/a≥2(achievedonlywhena=b)(4)Ifa,b∈R,thenab≤[(a+b)/2]^2≤(a^2+b^2)/2(5)Ifa,b∈R*,then1/2[(a^2+b^2)/(a+b)]≤ab≤(a+b)/2≤√[(a^2+b^2)/2]Specialnote:Intheaboveinequalities,"="isonlyachievedwhena=b.6.Cauchyinequality(1)Ifa,b,c,d∈R,then(a^2+b^2)(c^2+d^2)≥(ac+bd)^2(2)Ifa1,a2,a3,b1,b2,b3∈R,then(a1^2+a2^2+a3^2)(b1^2+b2^2+b3^2)≥(a1b1+a2b2+a3b3)^2(3)Ifa1,a2,...,anandb1,b2,...,bnaretwosetsofrealnumbers,then(a1^2+a2^2+...+an^2)(b1^2+b2^2+...+bn^2)≥(a1b1+a2b2+...+anbn)^2.II.AnalysisofQuestionTypesType1:Provinginequalitiesusingbasicinequality1.Giventhata,barepositivenumbers,provethatab≥(a+b)^2/42.Giventhata,b,carethreedistinctrealnumbers,provethata^2+b^2+c^2>ab+bc+ca3.Giventhata+b+c=1,provethata^2+b^2+c^2≥1/34.Giventhata,b,carepositivenumbersanda+b+c=1,provethat(1-a)(1-b)(1-c)≥8abc5.Giventhata,b,carepositivenumbersanda+b+c=1,provethat1/2[(a^2+b^2)/(a+b)]≤ab≤(a+b)/2≤√[(a^2+b^2)/2].1.設(shè)$a,b,c$均為正數(shù)，且$a+b+c=1$，證明：$$\frac{a}{b^2+3}+\frac{c^2+3}+\frac{c}{a^2+3}\geq\frac{1}{4}$$2.已知$a,b,c$為正數(shù)，且$a+b+c=3$，求證：$$\frac{a^2}{a^2+2b^3}+\frac{b^2}{b^2+2c^3}+\frac{c^2}{c^2+2a^3}\geq1$$3.求函數(shù)$y=\frac{x^2+1}{x+1}$的最小值。4.已知$x,y$為正數(shù)，且$x+y=1$，求證：$$\frac{x^3}{y^2}+\frac{y^3}{x^2}\geq\frac{1}{2}$$5.已知$a,b,c$為正數(shù)，且$a+b+c=3$，求證：$$\frac{a^2}+\frac{b^2}{c}+\frac{c^2}{a}\geq3$$1.對(duì)于正數(shù)$a,b,c$且$a+b+c=1$，證明不等式$$\frac{a}{b^2+3}+\frac{c^2+3}+\frac{c}{a^2+3}\geq\frac{1}{4}$$2.已知正數(shù)$a,b,c$且$a+b+c=3$，證明不等式$$\frac{a^2}{a^2+2b^3}+\frac{b^2}{b^2+2c^3}+\frac{c^2}{c^2+2a^3}\geq1$$3.求函數(shù)$y=\frac{x^2+1}{x+1}$的最小值。4.已知正數(shù)$x,y$且$x+y=1$，證明不等式$$\frac{x^3}{y^2}+\frac{y^3}{x^2}\geq\frac{1}{2}$$5.已知正數(shù)$a,b,c$且$a+b+c=3$，證明不等式$$\frac{a^2}+\frac{b^2}{c}+\frac{c^2}{a}\geq3$$已知正項(xiàng)等比數(shù)列$\{a_n\}$滿足$a_7=a_6+2a_5$，若存在兩項(xiàng)$a_m,a_n$，使得$a_ma_n=4a_1$，求$14m+n$的最小值。改寫：已知正項(xiàng)等比數(shù)列$\{a_n\}$，滿足$a_7=a_6+2a_5$。假設(shè)存在兩項(xiàng)$a_m,a_n$，使得$a_ma_n=4a_1$，求$14m+n$的最小值。2、求函數(shù)$y=\dfrac{x^2+7x+10}{x+1}$（$x\neq-1$）的值域。改寫：求函數(shù)$y=\dfrac{x^2+7x+10}{x+1}$（$x\neq-1$）的值域。3、求函數(shù)$y=\dfrac{x+2}{2x+5}$的最大值。（提示：換元法）改寫：求函數(shù)$y=\dfrac{x+2}{2x+5}$的最大值。4、已知$x,y>0$，滿足$x+2y+2xy=8$，求$x+2y$的最小值。改寫：已知$x,y>0$，滿足$x+2y+2xy=8$，求$x+2y$的最小值。5、已知$a,b>0$，滿足$ab=a+b+3$，求$ab$的范圍。改寫：已知$a,b>0$，滿足$ab=a+b+3$，求$ab$的范圍。6、已知$x,y>0$，滿足$x^2+y^2+xy=1$，求$xy$的最大值。改寫：已知$x,y>0$，滿足$x^2+y^2+xy=1$，求$xy$的最大值。7、已知$x,y,z>0$，滿足$x^2-3xy+4y^2-z=0$，求$\dfrac{xy}{z}$取得最大值時(shí)，$2x+y-z$的最大值。改寫：已知$x,y,z>0$，滿足$x^2-3xy+4y^2-z=0$。求$\dfrac{xy}{z}$取得最大值時(shí)，$2x+y-z$的最大值。1、若$a,b,c,d\inR$，則有$(a^2+b^2)(c^2+d^2)\geq(ac+bd)^2$，當(dāng)且僅當(dāng)$a/c=b/d$或者$ad=bc$時(shí)等號(hào)成立。2、二維形式的柯西不等式的變式：(1)若$a,b,c,d\inR$，則有$a^2+b^2\cdotc^2+d^2\geqac+bd$，當(dāng)且僅當(dāng)$a/c=b/d$或者$ad=bc$時(shí)等號(hào)成立。(2)若$a,b,c,d\inR$，則有$a^2+b^2\cdotc^2+d^2\geqac+bd$，當(dāng)且僅當(dāng)$ab/c=d$或者$ad=bc$時(shí)等號(hào)成立。(3)若$a,b,c,d\inR$，則有$(a+b)(c+d)\geq(ac+bd)^2$，當(dāng)且僅當(dāng)$ab/c=d$或者$ad=bc$時(shí)等號(hào)成立。3、二維形式的柯西不等式的向量形式：對(duì)于任意向量$\alpha,\beta$，有$\alpha\cdot\beta\leq|\alpha||\beta|$，當(dāng)且僅當(dāng)$\beta=k\alpha$或者$\alpha,\beta$共線時(shí)等號(hào)成立。4、三維柯西不等式：若$a_1,a_2,a_3,b_1,b_2,b_3\inR$，則有$$(a_1^2+a_2^2+a_3^2)(b_1^2+b_2^2+b_3^2)\geq(a_1b_1+a_2b_2+a_3b_3)^2$$當(dāng)且僅當(dāng)$b_i=ka_i(i=1,2,3)$時(shí)等號(hào)成立。5、一般$n$維柯西不等式：設(shè)$a_1,a_2,\dots,a_n,b_1,b_2,\dots,b_n$為任意實(shí)數(shù)，則有$$(a_1^2+a_2^2+\dots+a_n^2)(b_1^2+b_2^2+\dots+b_n^2)\geq(a_1b_1+a_2b_2+\dots+a_nb_n)^2$$當(dāng)且僅當(dāng)$b_i=ka_i(i=1,2,\dots,n)$時(shí)等號(hào)成立。題型分析：題型一：利用柯西不等式一般形式求最值。1、設(shè)$x,y,z\inR$，若$x^2+y^2+z^2=4$，則$x-2y+2z$的最小值為$-6$時(shí)，$(x,y,z)=(-2/\sqrt{3},1/\sqrt{3},-2/\sqrt{3})$。分析：由柯西不等式有$(x-2y+2z)^2\leq(x^2+y^2+z^2)(1^2+(-2)^2+2^2)=36$，即$x-2y+2z\geq-6$，等號(hào)成立時(shí)$(x,y,z)$與$(1,-2,2)$共線，即$x=-2/\sqrt{3}