高中数学竞赛讲义---不等式习题_SOS方法

不等式SOS方法1．设a、b、c为正实数，则222aabcbc证：左-右222aabcbc222222()()abacabacbcbc22()()()()ababcacabcbc222211()(()()()()ababbcbccaca222222()()0()()()()ababaabbcbccaca2．设a、b、c＞0，求证：3221223aaaabb①证：①322()02233aaabaabb2222()022baabaabb②当,,(0,2)abcbca时，②已成立。下设,,(0,2)abcbca，设max,,aabcⅠ。abcⅰ）2ab时，∵322222aaaabb，32223bbaabc(2)()0bbcbc也成立，∴32222233aababcaabbⅱ）2bc，类似可得32222323aababcaabb.当acb，则此时必有2ab.先证：32222aaabc39bc322227530cbcbcb增量易得再证：32229223accacccaa3226191420aacac.易得则：3322222239223aabccabcaabbccaa3．设,,abcR，求证：2222113aabcabab。证：原不等式222222111(1)3aabcababab2222222222()1()()112()()()23ababcbcabcabacabab22222222223()1()()()()()()ababcbcbcabcabacab222222223()1()()()ababcabcabacab42222222223()()(3)()abcabcababcabcbc42242222222223()()()33()abcaabcbcbcabcbcababcbc……①Ⅰ,,amaxabc,①244242222222222332()()3()0222aabcabcaabaacbcbcabbcbcbcbcⅡmin,,aabc，①中左233()3()bcababcbcabcⅢ(a-b)(a-c)＜0不妨设bac①中2243934acaac，23333bcababcab.∴原不等式成立4．设a、b、cR，证明：33321()aabcabc①证：33222()()(2)abcabcabcbcabac()abcA.①32aabcaA32()0aabcaA②33322222122aabcaaabcaabacabcabacAA()()ababacacAA∴②()()0ababacacA11()()0ababAB23()0cababAB.显然成立.5．求证：33331()bcabcabc①证：33222()()(2)()abcabcabcabacbcabcA,B、C类似定义，33331[222()]0bcabcabcabcbcAAA②而33222()bcabcbcA33222222()()()(2)bcabcbcbcbcabcabac3322222()()2()()(2)bcbcbcabcabcbcbcabac2()()(2)()bcbcabacbcbca∴②2()()1(2)0bcbcbcaabacbcAAA③注意到2abacbc2aA∴(2)bcaabacbcA2()(1)abcaA222aabcaaaAA22[]aaAaababaBAA22()3()()abcacabbcabaABAB2222()()3(22)abccabaacacbabABAB22()(222)cabacacbabAB2222222()()()()()()cabcabcabcacbaabbABABAB∴③22222()()()()()0bcBCbcCcbaBbcaAabbccABC则只须证：2222()()()()BCbcCcbaBbcaAabbcc0223223322332()44)2()()()0bcbcaCbcbcbcabcbcabcbc224222223333()()()3()()()()02222abcbcbcabcbcbcabcabcbcaa2332332243()()()()()2222bcbcbcbcbcbcaa2222433()()()()()bcbcbcbcabca2233()()()bcbcbc.显然成立.故原不等式成立.