备战数学应考能力大提升典型例题例1设,0xy,求证:211()()24xyxyxyyx证明:左-右=122xyxyxyxy221211022xyxyxyxyxyxy例2设a,b,c为正实数,求证:33311123abc+abc≥.证明:因为,,abc为正实数,由平均不等式可得33333331111113abcabc即3331113abcabc所以3331113abcabcabcabc,而33223abcabcabcabc所以33311123abc+abc≥例3设,xyR,1xy,求证:1125()()4xyxy证明:11254xyxy222222222251042512104332041804124xyxyxyxyxyxyxyxyxyxyxyxy1804xyxy成立例4已知01a,01b,01c,求证:(1)ab,(1)bc,(1)ca中至少有一个小于等于14.证明:假设1111,1,1444abbcca则有31112abbcca〔*〕又∵111abbcca11132222abbcca与〔*〕矛盾创新题型1.证明不等式nn2131211(n∈N*)2.设1223......(1)nSnn,求证:(1)(2)22nnnnnS()nN参考答案1.证明:(1)当n等于1时,不等式左端等于1,右端等于2,所以不等式成立;(2)假设n=k(k≥1)时,不等式成立,即1+k13121<2k,,1211)1(11)1(21121131211kkkkkkkkkk则∴当n=k+1时,不等式成立.综合(1)、(2)得:当n∈N*时,都有1+n13121<2n.