不等式的秘密（第一章）基本不等式

基本不等式ThebasicInequalities均值不等式AM-GMInequalityAM-GM不等式及其應(yīng)用AM-GMInequalityandApplicationsAM-GM不等式the:AM-GM不等對所有正實(shí)數(shù)a1,a當(dāng)且僅當(dāng)a1=a2=?=Forallpositiverealnumbersa1aEqualityoccursifandonlyifa1Proof.當(dāng)n=2時(shí),不等式顯然成立.如果不等式對n個(gè)正實(shí)數(shù)成立,那么它對2n個(gè)a因此不等式對n是2的指數(shù)冪形式個(gè)正實(shí)數(shù)是成立的.假設(shè)不等式對n成立,我們設(shè)a依據(jù)歸納假設(shè),我們有s因此,如果不等式對n個(gè)正實(shí)數(shù)成立,那么它對n-1個(gè)正實(shí)數(shù)成立,由歸納法(Cauchy歸納)可知,不等式對每一個(gè)自然數(shù)n都是成立的.當(dāng)且僅當(dāng)a1=a2=?=Proof.Theinequalityisclearlytrueforn=2.IfitistruefornaThustheinequalityistrueforeverynumbernthatisanexponentof2.Supposethattheinequalityistruefornnumbers.WethenchooseaAccordingtotheinductivehypothesis,wegetsThereforeiftheinequalityistruefornnumbers,itwillbetrueforn-1numbers.Byinduction(Cauchyinduction),theinequalityistrueforeverynaturalnumbern.EqualityoccursifandonlyifAM-GM不等式作為一個(gè)著名的、應(yīng)用廣泛的定理,在證明不等式方面也是不可缺少的.下面通過一些著名的不等式來研究它的強(qiáng)大的應(yīng)用.Asamatteroffact,theAM-GMinequalityisthemostfamousandwide-appbedtheorem.Itisalsoindispensableinprovinginequalities.Consideritsstrongapplicationsthroughthefollowingfamousinequalities.Nesbitt不等式pro:Nesbitt不等式(1)證明對所有非負(fù)實(shí)數(shù)a,(2)證明對所有非負(fù)實(shí)數(shù)a,Proof.(1)考慮下列表達(dá)式SMN我們有M+N=3,根據(jù)MN所以M+N+2S(2)考慮下列表達(dá)式SMN則M+N=4.根據(jù)MN因此,M+N+2S≥8,即2S≥4?S≥2其他證法Proof1.Usingthewell-knowninequalityxWegetaTherefore,abProof2.Settingxa(1)canbewrittenasyThisinequalityisclearlytruebythewell-knowninequalitypq+Proof3.(1)isequivalentto2Theinequalityfollowsbythewell-knowninequalityx3+Proof4.UsingtheAM-GMinequality,weobtainaSimilarly,wegetb2cAddingthreeaboveinequalities,wehaveaProof5.ABCthenBAABytheAM-GMinequality,wehaveAAThus,2A+BProof6.BytheAM-GMinequality,wehaveaaAddingtwoinequalities,weget2Thus,aSimilarly,wehavebcAddingthreeaboveinequalities,weobtainAddingthreeaboveinequalities,weobtainaProof7.Usingthewell-knowninequalityxWehave2Hence,4Similarly,wegetbcAddingthreeinequalities.wehaveaProof8.(ByCaoMinhQuang)UsingtheAM-GMinequality,wehave2Thus,aSimilarly,wegetbcAddingthreeinequalities,wehaveaWehavetoshowthat3Wesetx=3BytheCauchy-Schwarzinequality,wehavexBytheChebyshev’sinequality,wehavexMultiplying(3)and(4)yields,weobtain3Proof9.(1)isequivalenttoaThelastinequalityisclearlytrue.Proof10.BytheCauchy-Schwarzinequality,wehaveaProof11.Withoutlossofgenerality,wecanassumethata≥bBytheChebyshev’sinequalityandtheAM-GMinequality,wehaveaProof12.Settingx=ab+cxThus,1BytheAM-GMinequality,weget1SinceA>0,henceProof13.Usingthesamesubstitutionasthe12thproof,andsettingft=t1+fProof14.(ByCaoMinhQuang)Firstly,westateandprovealemma.Lemma.Ifxi,yi,ixwherei1,Proof(5).Wesetz1xItiseasytoseethaty1≥zTherefore,xLetusnowprove(1).Withoutlossofgenerality,wecanassumethata≥b≥abaAddingtwoinequalities,weobtain2aaProof15.Withoutlossofgenerality,wecanassumethata≥b≥c(1)becomesxUsingtheAM-GMinequality,wehavexItsufficestoprovethat2ThelastinequalityitruesincexToprove(6),besidetheaboveproof,wealsohaveananotherproof.Proof16.Wesetm=m

Wenotethat7m-24Thisinequalityisclearlytrue.Proof17.(ByCaoMinhQuang)Bysettingx=a(1)canbewrittenasxUsingtheAM-GMinequality,wehavexUsingtheAM-GMinequalityagain,wehavexAddingtwoinequalities,weget7Since(1)inthehomogeneous,wecanassumethata+aWehavesomeproofsof1Proof18.For0<4Hence,4x≥1aProof19.(ByCaoMinhQuang)For0<3Hence11-Therefore,byCauchy-SchwarzInequality,wehaveaProof20.(ByCaoMinhQuang)For0<2Therefore,xUsingthewell-knowninequalityxWegetaProof21.(ByCaoMinhQuang)BytheAM-GMinequality,weaSimilarly,wehavebAddingthreeinequalitiesandnotingthataWeobtainaProof22.Settingft=t1-t?Proof23.Withoutlossofgenerality,wecanassumethata≥b≥c.Sinceft=tforaWearedone.Proof24.(ByCaoMinhQuang)Toprove(1’),weneedtoprovealemma.Lemma.(CaoMinhQuangandTranTuanAnh)Ifa,b,aProof.Usingthewell-knowninequalityxWegetaWenowprove(1’).Usingthewell-knowninequalityxyWeobtainaLast,weuse"MixingVariablesTheorem"toprove(1).Proof25.BysettingEa,b,cEThus,Ea加權(quán)AMGM不等式pro:加權(quán)AMGM不等式假設(shè)a1,a2,?,an是正實(shí)數(shù),如果n個(gè)aProof.這個(gè)不等式的證明和經(jīng)典AM-GM不等式的證明是類似的.在n=2的情形,我們必須詳細(xì)地證明(因?yàn)椴坏仁街谐霈F(xiàn)了實(shí)數(shù)指數(shù)).我們先來證明,如果x,y≥0ax證明這個(gè)不等式的最簡單的方法是考慮x,y是有理數(shù)的情況,至于實(shí)數(shù)我們可以采用極限的方法來進(jìn)行.如果x,y是有理數(shù),設(shè)x=mm+ma如果x,y是實(shí)數(shù),則存在兩個(gè)有理數(shù)序列rr于是a或者a取極限,令n→+∞,ax

?盡管AM-GM不等式非常簡單,但它在數(shù)學(xué)競賽中的不等式證明方面扮演著重要的角色.用下面的一些例子來幫助你熟悉這個(gè)重要的不等式.設(shè)a,b,c>0aProof.注意到恒等式2則該不等式等價(jià)于cyc由AM-GM不等式,我們有cyc所以,不等式成立.

?(IMOShortlist1998)設(shè)x,y,z>0xProof.利用AM-GM不等式,我們有x因此cyc當(dāng)x=y=z=(APMO1998)設(shè)x,y1Proof.不等式整理之后等價(jià)于x由AM-GM不等式,我們有3即x

?設(shè)a,b16Proof.應(yīng)用AM-GM不等式,我們得到16當(dāng)且僅當(dāng)a=b=c=d(PhamKimHung)設(shè)a,b,c是周長為1Proof.設(shè)x=b+c不等式變?yōu)?又設(shè)m=xym由AM-GM不等式,我們有m兩式相乘即得m所以,原不等式成立.

?(PhanThanhNam)設(shè)a1,a2,?,an是正實(shí)數(shù),2Proof.根據(jù)AM-GM不等式,我們有a將上述不等式對于k∈{1,2k這里的指數(shù)cic這是由于ai≤a當(dāng)且僅當(dāng)ai=ii=(USAMO1998)設(shè)a,b,c1Proof.注意到a3+abc類似地可得另外兩個(gè)不等式,并將它們相加,我們有abc所以,原不等式成立.

?注意IMOShortlist1996類似的問題.設(shè)x,y,z是積為xy如果x1,x2,?,證明xProof.所給條件變形為1應(yīng)用AM-GM不等式,我們有x類似地可得其他n-1個(gè)不等式,并將它們相乘,設(shè)x,y,z是正數(shù),且滿足xProof.注意到x從這個(gè)形式,我們利用AM-GM不等式10類似地,可得10將上述三個(gè)不等式相加,我們有10于是,只需證明下列不等式x而這是成立的.事實(shí)上x

?(MathlinksContest)設(shè)a,baProof.由AM-GM不等式,我們有LHS≥于是,我們只需證明a由于abc=1,所以式*ab根據(jù)AM-GM不等式,我們有2由此4所以,原不等式成立.當(dāng)且僅當(dāng)a=b=設(shè)a,b,caProof.由加權(quán)AM-GM不等式,我們有a整理即得a當(dāng)且僅當(dāng)a=b=c時(shí)設(shè)a,baProof.由恒等式ab我們有ab利用AM-GM不等式,以及a我們有2即ab由式1,2a當(dāng)且僅當(dāng)a=b=1,c設(shè)a,b1Proof.注意到ac根據(jù)AM-GM不等式,我們得到ac此外cyc當(dāng)且僅當(dāng)a=b=c=設(shè)a,b,c,dabcProof.不失一般性,我們設(shè)e根據(jù)AM-GM不等式,我們有abc因此,只需證明e這是成立的,可由e-12e(PhamKimHung)設(shè)a,b1Proof.我們來證明1凷AM-GM不等式2將上述不等式相加,即得所要證明的不等式.

?注意(1)使用類似的方法,我們可以證明5個(gè)變量的類似的不等式.為此,我們有a(2)本題不等式加強(qiáng)為1(3)猜想:設(shè)a1,1(IMO2006)求不等式ab對所有實(shí)數(shù)a,b,c都成立的最小的記x=a-b其中,s,x,y,z事實(shí)上,s是一個(gè)獨(dú)立變量.首先我們來考察xyzz和x2+y2+z2之間的關(guān)系.由于x+y+z=0,很顯然,x,y,z中有兩個(gè)變量的符號相同sxyz等號當(dāng)x=y時(shí)成立.設(shè)t=x+y2于是4結(jié)合式1,2sxyz這就意味著M≥9232.為證明M=9232是最好的常數(shù),我們必須求出sa使用AM-GM不等式的重要的原則是選擇合適的系數(shù)滿足等號成立的條件.例如,在例1.1.2中,使用下列形式的AM-GM不等式是錯(cuò)誤的(因?yàn)榈忍柌怀闪ⅲ﹛對于每個(gè)問題,為AM-GM不等式給出一個(gè)固定的形式是很困難的.這取決于你的智慧,但是尋找等號成立的條件是可以幫助我們做到這一點(diǎn)的.例如，在上面的問題中,猜測到等號成立的條件是x=y=z=1x變量相等使等號成立的這樣的問題,在使用AM-GM不等式之前是很容易確定的.對非對稱問題,這個(gè)方法需要有一定的靈活性(參見例1.1.13,1.1.14和1.1.16).有時(shí)你需要建立方程組并求出等號成立的條件(這個(gè)方法稱為“平衡系數(shù)法”,這部分內(nèi)容將在第6章中討論).Cauchy求反技術(shù)在本節(jié)中,我們將AM-GM不等式關(guān)聯(lián)到一個(gè)特別的技術(shù),稱為Cauchy求反技術(shù).出乎意料的簡單,而且十分有效,是這一技術(shù)的特殊優(yōu)勢.下面的例子體現(xiàn)了這種優(yōu)勢.(BulgariaTST2003)設(shè)a,baProof.事實(shí)上,直接對分母使用AM-GM不等式是不行的,因?yàn)椴坏仁礁淖兞朔较?即a但是,我們可以以另外的形式使用AM-GM不等式a這樣,不等式變成ccc這是由于3cy?ab≤ecc?注意?使用類似的方法可以證明下列結(jié)果.設(shè)a,b,c,d是正實(shí)數(shù)a這個(gè)解法似乎像變魔術(shù),AM-GM不等式應(yīng)用