2.3.2放缩法课后导练基础达标1设x>0,y>0,A=yxyx1,B=yyxx11,则A,B的大小关系是_________________.解析:A=yxyyxx11
log4564=53,∴log54234b>c,则比较大小:))((cbba_______2ca.解析:∵a>b>c,∴a-b>0,b-c>0,22)()())((cacbbacbba.∴))((cbba≤2ca.答案:≤5求证:2<2log3log32<2+1.1证明:2log3log322log3log232=2,又2log3log32<23log4log32+1,∴2<22log3log32+1.综合应用6已知a>b≥0,c>0,求证:bcbaca.证明:bcbcbcbcbcbcacacaca)(.7求证:2(1n-1)<1+nn213121(n∈N*).证明:∵)1(2121kkkkk,∴1+)23(2)12(213121n)11(2)1(2nnn又),1(2121kkkkk∴nnnn2)1(2)12(2)01(213121∴原不等式成立.8已知an=)1(433221nn(n∈N*),求证:2)1(nnn,∴an=2)1(21)1(3221nnnnn.又)1(nn<21［(n+1)+n］=21(2n+1),2∴an=2)1(2122523)1(32212nnnn∴2)1(2)1(2nannn.9求证:91+41)12(14912512n(n∈N*).证明:)111(41)22121(21)12(12nnnnn,∴左式<41［(1-21)+(21-31)+…+(111nn)］=41(1-11n)<41.拓展探究10在△ABC中,求证:3≤2cbacCbBaA(a,b,c为三边,A,B,C为弧度).证明:∵b+c>a,有a+b+c>2a,∴,21cbaa可知2AcbaaA.同理,2,2CcbacCBbbabB.∴22CBAcbacCbBaA.又∵(a-b)(A-B)≥0,便是aA+bB≥aB+bA,∴aA+bB+cC≥aB+bA+cC.同理,aA+bB+cC≥cA+aC+bB,aA+bB+cC≥cB+bC+aA.三式相加,得3(aA+bB+cC)≥π(a+b+c),即cbacCbBaA≥3.∴原不等式成立.备选习题11设n∈N,且n>1,f(n)=1+21+31+…+n1,求证:f(2n)>22n.证明:f(2n)=1+21+31+…+n1+…+n21=1+21+(31+41)+()21221121(8171615111nnn22)212121()81818181()4141(211nnnn12设三角形三边a,b,c满足关系an+bn=cn(n≥3,n∈N),求证:△ABC为锐角三角形.证明:∵an+bn=cn,3故(ca)n+(cb)n=1.∴c>a,c>b,△ABC中c边最长.又由于n≥3,1=(ca)n+(cb)n<(ca)2+(cb)2,∴a2+b2>c2,由余弦定理cosC=abcba2222>0,△ABC为锐角三角形.13设a1,a2,a3,…,an是一组正数,求证:12212321322121)()()(aaaaaaaaaaaann证明:,11)()(21121122212aaaaaaaaaa,321213212132321311))(()(aaaaaaaaaaaaaaa，))(()(21121221nnnnnaaaaaaaaaaannaaaaaa2112111∴232132212)()(aaaaaaa.111)(1211221aaaaaaaaannn14α≠2n(n∈Z),求证:(1+n2sin1)(1+n2cos1)≥(1+2n)2.证明:左式=1+n2sin1+n2cos1+nn22cossin12|cossin|1|cossin|21nnnn≥(1+2n)2(∵|sinα·cosα|≤21).∴原不等式成立.15设0<α<2,0<β<2,0
