竞赛数学(张同君陈传理)代数3(不等式)

代数不等式解不等式在实数范围内,用不等号“＞(或≥)”,“＜(或≤)”,连结的两个解析式子，叫做不等式。用“＞”或“＜”连结的，叫做严格不等式，用“≥”或“≤”连结的，叫做非严格不等式。使不等式成立的自变量的所有值的集合，称为不等式的解集。解集相同的两个不等式，称为同解不等式（或等价不等式）。不等式的性质1;2,;3;i,;ii,;4,0;abbaabbcacabacbcabcdacbdabcdacbdabcacbd反对称性：传递性：加法单调性（保序性）：由此可推出：（同向可加性）（异向可减）不等式的性质5,0;i0,00;11ii00;iii0,00;60,1;70,1.nnnnabcacbcabcdacbdababababcdcdabnnNababnnNab由此可推出：且且例题1591.xx例：解不等式50,5,9.90.519.51929.21529.xxxxxxxxxxx首先确定允许值范围：其次，原不等式变形为等价不等式两边显然都是非负的，因此可以两边乘方，得到不等式所以21512150,,.215,.,5,2;1522150,,,2,21549.15,9,,2147|92xxxxxxxx若即时不等式左边为负值而不等式右边为非负值以上不等式无解因此原不等式在[)上无解若即时不等式左边、右边均为非负可进一步乘方将其转化为等价不等式在上解以上不等式即得到原不等式的解集是.例题25231.xx例：解不等式3330,5,,5,5.22215,55,2323,5231,9.55.325,55,2323,52131231,,55.32333,55,23232xxxxxxxxxxxxxxxxxxxxxxxxxxxx取点分别在区间上讨论当时有原不等式变为解得与的条件结合得当时原不等式变为解得与结合得当时,53231,7.,7.211,2,35231|7.3xxxxxxxxxx原不等式变为解得与条件结合得综合可得不等式的解集为或例题2225162423log1.14xxx例：解不等式222222,2511,16251,1624225.1416,9,33,31.171.16170.xxxxxxxxxxx由对数函数的性质知当时原不等式等价于不等式组解不等式组得222222225201,162501,16242250.1416,925,5335,2240,64,34.117.16170.12xxxxxxxxxxxxxxxx当时原不等式等价于不等式组解不等式组得或综合、得原不等式的解集为|31|34.xxxx例题46964136.xxx例：解不等式11111111212110,6964136.3260,6,6613.23366136,0,6130,0.2232330,61360.,,3232233322xxxxxxxxxxxttttttttttt显然不等式可化为因为所以两边同除以得令则不等式变为即因为所以解得即亦即.211,31,11,121.10,1,110,1.1|,11,.xxyxxxxxxxxxxxx而是增函数所以即当时得当时得所以原不等式的解集为练习2112211221.log2log11.2.3loglog1.xxxxx211222221.log2log1121020101022230223.03|23.xxxxxxxxxxxxxxxxxxx故原不等式的解集为1122121122212112211222.3loglog103log00log103log0log103loglog100111.log11log22421|.4xxxxxxxxxxxxxxxxxx或解得或故原不等式的解集为证明不等式在定义域中恒成立的不等式叫做恒不等式。确认一个不等式为恒不等式的过程，为证明不等式。证明不等式的主要方法是根据不等式的性质和已有的恒不等式，进行合乎逻辑的等价变换。主要方法有：比较法、综合法、分析法、反证法、数学归纳法、放缩法、构造法和导数法。例题35,,,:23.23abcababcababc例：设为正数证明33363323322332222232323232.,,0,0.32322222220..2abcababcababcababcabcabcabxabycxycabcabyxyxyxyxyxyyxyxxyxyxyxyxyxyxyxyxyxxyabcab设则仅当即时等号成立所以33.23abcababc例题3610,:;2,,0,:;30,:11;401,:arcsincoscosarcsin.aabbabcabcabeeeeabcabcabcabaabb例设求证求证设求证已知求证011111,0,0,,1.10,.aabbababababababababababaabbeeeeeeeeeeeeeeeeababeeeeeeeeeee作差因为所以所以所以322233333333333332,,,0,,.0,1,abcabcabcbcacababcabcabbcacabcabcabcabcabcabcabcabcabcabcabcabaabcbccababcbc由于的对称性所以不妨设与均为正数作商由于所以33331,10,0,0.1111.abbcacabcabcaabbcaccababccabcabc且所以所以30,10,10.111111110,110,1111.abaabbaabbaabbababaabbaabb因为所以作倒数差所以2222222222401,arcsincosarcsinsin0,22cosarcsin1sinarcsin10.121,24=210,448因为所以作平方差所以21,arcsincoscosarcsin.2因此例题80117:1617.kk例证明80180211121,,111,12112121121211,.180,116,180,2128011281117.nkmkkkkkkkkkkkNkkkkkkkkkknmnmmnmknmknmk由得故即得从而为自然数取得取得因此80111617.kk例题8,,,:.1111abcRabcabcabcabc例求证122121212112,[0,)0,10.1111[0,),1.11111.111xfxxxxxxxxxfxfxxxxxxfxxabcabcfabcfabcabcabcabcabcabcabcabcabcabcabc构造函数则当时所以函数在上是严格递增的由有即例题29,21121.212xxx例证明对任意实数均有2221,110,0,,1410,4410.21212xfxyxyxxyxyxyyyyy设则有将上式看作关于的一元二次方程此时必为实数则即解此不等式得2.2,0,,21121.212yyxx显然当时也满足上式所以成立例题111101,:231;2221.2nfnnNnnmfnfmnmnnfn例已知求证11,111111111123123111111.12nmnnmfnfmmmnmnmmmnnnnn个当时有2111112.1112542,21.2341222,2,1,21112221222221112222122221,.22222,212kkkkkkkkkkkknnfknkfnkffkkkknfn用数学归纳法证当时成立假设时命题成立即则当时结论成立综上不等式成.立例题12,,:.lnln2ababababab例设是不同的正数证明220,21122lnln1lnlnln.1,211ln,1.1211ln,.111,baaaababababbabaabababbabababbaxbxxxxxxxxfxxfxxxxxfx不妨设原不等式转化为令则所证不等式等价于其中考虑函数其导数为当时0,[1,).2110,1,0,1,ln1..lnln1ln..fxxfxfxxxxabababxxx故在区间上为增函数又所以当时即当时成立故成立类似可证成立于是原不等式成立例题222222130,0,0.:.xyzxxyyyyzzzzxx例设求证余弦定理2222coscababCABCabc222222,,60,,,.,,.,,.OOABCAOBBOCCOAOAxOByOCzABxxyyBCyyzzCAzzxxABCABBCAC