hojoo lee代数不等式100题.pdf
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1、222222100,-11:,0,1,:113()1:11235:19hojoo leeaba babproveabababsolution oneababasolution twoa代数不等式第一章的题 分享下自己的做法 水平有限不足之处请指正 天书由柯西注意到2221521(21)0,:()911993abaasoabab2222222222222:,:3(1)139933346()33346()()()4433333333(2)(2)()022442224x yR provexyxyLHSRHSxxyyxyxxyyxyxyxyxyxyxyxyxy 13:0,1,:1:1(1)(1):1y
2、xyyyxx yprove xyxxxxxxyxyxyxyxxyso xyxyxy由伯努利32333322223333222223333223322422466334:,0,:2432()()()1):4442):43334334abaa babbaabbaba bproveaa babbabababaa babbaabba babababx yx yxyx y由柯西4224336633222332222222222222222223332233623()()()()()()30:()93,.3):32x yx yx yxyx yx yxyxyxxyyxyxyxxyyx yxxyyx yx y
3、doneaabbab由均值332233224224333322422432233982416162727aabba bababa ba baba bababa ba b 1212844866933912126684486693396633233244222333322223324423322227272416162272754161632()()27()16()()()27()16()(xyx yx yx yx yx yxyx yx yx yx yx yx yx yxyxyx yxyx yxyxxyyxyx yxyx yxxyyx22233233222332224422223323322222
4、223323342243322233233222)()4()12()27()()()12()27()()()12181233()()12()xyyxyx yxxyyx yxxyyx yxyxxyyxyx yxxyyxy xyxxyyxyx yxx yyx yxyxxyyxyx yxy222224332222243322224332233223()()()12()3()()9219:()27()9219,.xxyyxyxxyyx yxyxxyyxxyyx yxxyyxxyyx yxxyyx yxxyydone注意到33333115:,0,:2()()113111122()():113111122
5、()()aba bproveababbaaabbababababbbaaabababab注意到两式相加整理得原不等式2222222226:0,:22()242222xyxyxyxyx yR provexyxyxyxyxyxyxyxyxyxyxyxyxy2222222222222224122222()2122()22xyxyxyxyxyxyxyxyxyxyxyxyxyxyxyxyxyxyxy 22222222222222222222222221()22()222222222222()()22()2222221()()22222xyxyxyxyxyxyxyxyxyxyxyxyxyxyxyxyxyx
6、yxyxyxyxyxyxyxyxyxyxyxyxyxyxyxyxyxyxy 02.2()0()21()xyxydonexyxyxyxyxyxy 若则证毕若333222222333222222222222222()9()7:,0,:33():.()()9()2()9()627()()9()()()0abcabca b cproveabcabcsolution one S O Sabc abababcabcabcabcabcabcabcabc abcababcabc2222222323229()0():()()9,:,266926183318226223322(3)(92)6(3)(392abca
7、babc abcabc abcabcsolution two pqrpabc qabbcca rabcppqrpppqqrrpqpqqqpqqqrqqrq注意到证毕做代换不妨设24222)(1846)01845445814:18460:661845814184:(3)(454)0186qrqp qpqqqqrrrpqqqqq下面只需证明由四次舒尔下面只需证显然成立3332338:,0,:3322,3333:,.333xyyzzxxyzx y zprovexyzxzxyzxyxyzxyzxyzzxyyxyzy zzdone不妨原不等式注意到33333339:,0,:()()()3()()():3
8、()()()a b c x y zproveax by czabcxyzabcabcaxbyczax by czxyzxyzaxbyczax by cz注意到两式相加整理得原不等式210:,0,:1()()()()()():1,11tansin22:tan;tan;tan11222tansecsin1222sinsin2sinsinsin1:cos22222xyzx y zprovexxy xzyyxyzzzx zyxxyyzzxxxAAABCxyzAAAABABC齐次的不妨原不等式再做代换展开得:注意到恒等式222cos122:coscoscos222BCBCABACBC故只需证明由柯西这是
9、显然的222222223311:,0,1,:21111:()(),:(1)(1)(1)11321()()13:(1)1xyzx y zxyzprovexyzsolution onef xf xxjensen xfxyfyzfzf xyzxyzxyzxyzxsolution twoxxxx构造减函数 凸函数由加权由赫尔德下面只需证明22221:(1)33xxxyz柯西显然211112:,1,2,:11111:11x y zprovexyzxyzxyzxxxxxxx 由条件知两边开根号得原不等式22222233333222313:,0,:()()()()()()()1(1)(1)(1):;1()(
10、)1(1)(1)(1)()a b cprovea bb cc a abbccaabcaabc babc cabcabcbcabccaabbcaabcabcabcxyzxyzbcazxyxyz xyyzzxxyzxy 原不等式做代换原不等式33311()()211(2)(1)0 xyxyttttt tt 不妨设原不等式这是显然的4442222223333334224223333422423342233622642244614:,0,:2()2:():a b cproveabca bb cc aa bb cc aabbccaaa baa ba baba babaa baa bca babaa ba
11、 baba ba ba b ca ba 两边平方由均值和四次舒尔下面只需证明23342223322226226662622644222221:()0211()02211()022bca b ca b ca b ca b caba ba ba bccaba ba ba ba bab注意到显然显然显然三式相加得证!4224222222222444422222222244444215:,0,:21:2222212()()22222222,.a b cproveaa bbaabca ba ba ba bLHSababa bc aa bc aa bcaaaaaaabc done注意到2222222222
12、222222223 216:,:(1)2:2(1)(1)(1)(1)(1)(1)(1)(111)(111)3 2a b cR proveabMinkowskvababbccabacbacabcabcabcabc 由222222222222222222222222217:,0,:,:2()()()()()()2222:a b cproveaabbbbccaaccsolution one Minkowskibaabbbbccaabbcbcbacabbcbbaaccaaccsolution twoaabbb不等式为了消去按照如下方式分组配方222222222222222222()()24()()2(
13、)0bccaaccaabbbbccabbaabbbbccabbabbcca显然成立222222222218:,02:()()()()()():(,),(,),(0,0):2:,a b c dadbcproveacbdabcdacbdacbdMinkowskiA a b Bcd OOABHOHABAOBOOHHAOA OHHBOB右边不等式显然成立左边 构造点记 到的垂足是那么原不等式的几何意义为由三角不等式相加证毕19:,1,:111(1):(1)(1 1)(1)11(1 1)(11)1a b cproveabcc abc abcabcabcaba 注意到322333()()()20:,0,:
14、83()8:()()()0993()()():,.8993ab bc caabbccaa b cprovecabab bc caaabababaabab bc caabbccasodone 注意到222111321:,0,:2111:3tan,tan,tancos2:111133:,1,221:(x y zxyzxyz provexyzsolution onexA yB zCAjensensolution twoacabcabcxyyzzxbbacacacacbacb三角代换做代换:原不等式由不等式显然成立均值做代换原不等式注意到1322)()acabbca bc223223222:,0,:1
15、8:(8)8:(8)aa b cproveabcaa abcaabcaa abc方法太多由赫尔德下面只需证明此为三次舒尔显然成立33331123:1,0,:6:11111136(16)916918333 3331311:183 33:16(16abbccaa b cprovebaabcsolution onebabbcabbcaabcabcabcabcabcabcsolution twobaa 均值下面只需证明由条件易知这是显然的赫尔德33333311111)(16)13:2713 3babaaaaa b cabc 下面只需证明由条件易知这是显然的24:,0,:()4()()()3:()()(
16、)()3,.a b cproveab ababcaba bc ab acabca bc ab aca bc b cabc done平方注意到25:,0,:22:2()abca b cprovebccaabaaabcabca bc注意到322222326:,0,:222636()0abca b cprovebccaabaa bababca bababcc ab方法太多显然成立223127:1,0,:(1)1:,(1)1()()3abca b cproveabxyzxzabcyzxyyxzyxyzx yxyxxyz 方法太多做代换原不等式此为三次舒尔显然成立11111128:,0,:1111113
17、111:1;1;1111:(1)(1)(1)5354,1a b cproveabcacbbcabacxaybzcbcaxyzabcaaxxyzxyabcaaa b c做代换则由抽屉原理知当中有俩在 的323232,:0,0,0:0443:044343,.3abxyzyzxycase one xyzxyxyzxycase two xyzxyxyztxytttdone 同侧 不妨设为 和那么和 当中必有一个大于等于不妨设设得到333329:,0,1,:(1)(1)4(1)(1)133:3(1)(1)884(1)(1)244xx y zxyzproveyzxyzxxxyzyz由均值555544332
18、230:1,0,:11:11ababca b cproveabababababccabcababa bb aababa bb aabc注意到322231331:1,0,:2()13:()22()abca b cprovea bcabb ca bcaba bc注意到2222222232:,0,:232323233:(23)():233()2(23)322()11:22;()()02()22abcda b c dprovebcdcdadababcaa bcdabcdbcdabcda bcdaabacbdaabacbdaacbd由柯西下面只需证明又有两式相加证毕3332133:,0,1,:3111:
19、31812633()111:1()()2,:,.4333aa b c dabbccddaprovebcdabcdaaabcdbcdabcdabbccddaac bdasoadone注意到又有2211122222211122212121222222222111222111222111222111222211221211111834:,0,:()()()11884()()2()()()()822()kkkxyzR x yzx yzprovex yzx yzxxyyzzx yzx yzx yzx yzx yzx yzx yzx yzx y x yz zx yz2222222121212112212(
20、)88()()()()()x yzxxyyzzx yx yzz222222235:,0,:122,:1,22abca b cproveabcabcaadoneabcaab只需证明其中一个 不妨证左边的由柯西333334436:,0,:44aa b cproveabcaabcaabcbccaab注意到两边求和即为原不等式3333221137:,0,:1111:a b cproveabcababccabcabcabcababca bb aabc注意到222222232232()338:,0,:5()4129335269131:526926911:512692692(26)(96)910bcaa b
21、 cprovebcaaaabcaaAxBaaxxxABAxAxAxBxBxBAxBA xAB xB 齐次的不妨设原不等式我们希望得到这样一个局部式时取等我们希望这个式子含222(1)132023:,2525123:2(1)(21)02525269xxABABsoxxxxx有因式故令是其导数的零点解得显然成立2222222222(2)39:,0,:82()69383696986868773693693(1)68686:7,.63(1)6abca b cproveabcaaabcaaaaaaaaaaaaadonea齐次的不妨原不等式注意到22223233333233311140:,0,:1()()
22、00()()11:()111bccaaba b cproveabcabcbcacabbcab acaabcaabcab acacabcbaabcaabcaabccabccabcbabcaabccabc不妨设注意到33()()11:,()()0ab bcab bccabcbabc由于故原式成立222241:,0,:()()()11111()()()()11,()()0,abca b cproveabcabcabcab acLHSRHSabcbcacbab bcacbcabacabbcabab bcbcac不妨由于故原式成立322223322222222222222142:,0,1,:322311
23、211()()0,.aa b cabcproveabcaaaabcbcabcabcaaaaadonebcbcbc原不等式222221343:3,0,:121(1 1 1)993:11 1 1323abca b cproveababababa 由柯西及均值22222222222222221944:,0,1,:1232312124():3122a b cabcproveabababababababababababcabcacbc原不等式注意到44422224442222222222222211145:,0,3,:14441100443:33434333(3)()04343a b cabcprove
24、abbccaaba babcababa babca bb cc aabbccaaba baca cbabcacabbc bcaca cbcb不妨原不等式注意到22222222222222222222222233(3)()0111:(1,1,1)(,)434343111:(3)0,.34343caa cbb cacbc aba ba cb caba baca cbcc ba bsoa bdoneaba baba b故和,同序222222222222222346:,0,1,:4111:(1)141:(1)33abca b cabcprovea ab bc cbcaaaba ababa b由柯西下面
25、只需证明由均值这是显然的22222222222111947:,0,()()4()()()94()3324()()()2()()4()21()0()0()()():21()()(cca b cabbccaabbccaabcababcabababababac bcababS abac bcababcSac bca记作不妨设2222222222222330)21()()()210()()():()()()()()()()1()()()()()()()abcababbSac bacbSab bcacbcsoS abSbcSacSbcS cabcbcbcacbcac baab bcbcbcacabbc3
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