(II)因为
f(x1)x1f(x2)x2g(x)?f(x)x上是增函数,所以
在(0,??)?f(x1?x2)x1?x2f(x1?x2)x1?x2?f(x1)?x1x1?x2x2?f(x1?x2)
??f(x2)?x1?x2?f(x1?x2)
两式相加后可以得到f(x1)? (3)
f(x1)x1?f(x1?x2???xn)x1?x2???xnf(x2)?f(x1?x2)
x1x1?x2???xn?f(x1?x2???xn)?f(x1)? ……
f(x2)x2f(xn)xn?f(x1?x2???xn)x1?x2???xn?f(x2)?x2x1?x2???xnxn?f(x1?x2???xn)?f(x1?x2???xn)x1?x2???xn?f(xn)?x1?x2???xn?f(x1?x2???xn)相加后可以得到:
f(x1)?f(x2)???f(xn)?f(x1?x2???xn)
所
令
xn?1以,有
2x1lnx1?x2lnx2?x3lnx3???xnlnxn?(x1?x2???xn)ln(x1?x2???xn)?11112222???22ln2?32ln3?42ln4???(n?1)2ln(n?1)???111?2??ln??2?2???3(n?1)??2????
(1?n)?????
?1111?2?2?2?2???34(n?1)?2?111???22?32???(n?1)2?
??1?11??ln?????????(n?1)n??2?13?2?
1??11?n??????????2(n?1)(n?2)?n?1??2n?2?122
n2(n?1)(n?2) 所以
ln2?2132ln3?2142ln4???21(n?1)ln42ln(n?1)?2(n?N).*
(方法二)ln(n?1)(n?1)22?ln(n?1)2(n?1)(n?2)?1??1?ln4???(n?1)(n?2)?n?1n?2?
所以
122ln2?2132ln3?2141222ln4???21?nln4?12ln(n?1)?ln4????(n?1)?2n?2?2(n?2)21 又ln4?1?1n?1,所以
ln2?2132ln3?2142ln4???21(n?1)2ln(n?1)?2n2(n?1)(n?2)(n?N).* 例16.(2008年福州市质检)已知函数f(x)?xlnx.若a?0,b?0,证明:f(a)?(a?b)ln2? 解析:设函数g(x)? ?f(x)?xlnx,?0?x?k.?g?(x)?lnx?1?ln(k?x)?1?ln令g?(x)?0,则有xk?x?1?xk?x,k2?x?k.f(x)?f(k?x),(k?0)f(a?b)?f(b).
?g(x)?xlnx?(k?x)ln(k?x),2x?kk?x?0? ∴函数
g(x)在[k2,k）上单调递增，在(0,k]上单调递减.
2kg()2 ∴g(x)的最小值为
，即总有g(x)?k g().2
6
而g(k)?2kkkf()?f(k?)?kln?k(lnk?ln2)?f(k)?kln2,222
?g(x)?f(k)?kln2,
即f(x)?f(k?x)?f(k)?kln2.
令x?a,k?x?b,则k?a?b.
?f(a)?f(b)?f(a?b)?(a?b)ln2.
?
f(a)?(a?b)ln2?f(a?b)?f(b).
三、分式放缩
姐妹不等式:ba?b?ma?m(b?a?0,m?0)和
ba?b?ma?m(a?b?0,m?0)
记忆口诀”小者小,大者大”
解释:看b,若b小,则不等号是小于号,反之. 例19. 姐妹不等式:(1?1)(1?1)(1?1)?(1?35(1?12)(1?14)(1?16)?(1?12n)?12n?112n?1b?ma?m1?(b?a?0,m?0)12n?1)?2n?1和
也可以表示成为
可得
2?4?6??2n1?3?5???(2n?1)?2n?1和1?3?5???(2n?1)?2?4?6???2na?解析: 利用假分数的一个性质b
?(2462n????3?5?7?2n?1?1352n?12462n352n?1???(2n?1)2462n1112462n2)????)?2n?1即(1?1)(1?)(1?)?(1?352n?11352n?12n?1.
例20.证明:(1?1)(1?)(1?)?(1?471113n?2)?33n?1.
解析: 运用两次次分式放缩:
2583n?13693n??????.?????1473n?22583n?1 (加1) (加2)
21?54?87???3n?13n?2?47103n?1.?????3693n 相乘,可以得到:
3n?1?47103n?11473n?2?258???????????(3n?1)????????.?3n?2?2583n?12583n?1?1472
所以有(1?1)(1?1)(1?1)?(1?4713n?2)?33n?1.
四、分类放缩
例21.求证:1?1?1???2312?1n?n2
?1?12?(14?14)?(123 解析:
(12n1?12?13???12?1n?123?123?123)???
?12n???12n)?12n?n2?(1?12n)?n2
7
例22.(2004年全国高中数学联赛加试改编) 在平面直角坐标系xoy中, y轴正半轴上的点列
?An?与曲线y?2x（
x≥0）上的点列?B?满足OAnn?OBn?1n，直线AnBn在x轴上的截距为an.
点Bn的横坐标为bn,n?N?.
(1)证明an>a>4，n?N; (2)证明有n0?N?，使得对?n?n都有b2?n?10b1??b3b2???bnbn?1?bn?1bn 解析:(1) 依题设有：A bn?2bn?2n?1?0,?n??,Bnbn,??2bn,?bn?0??，由OB1n得： n1n2,?bn?1n2?1?1,n?N*,又直线AB在x轴上的截距为an满足 nn?an??0???2bn?1??1????0???bn?0?n??n? an?bn1?n2bn ?2n2bn?1?nbn?0,bn?2?221nbn2 ?an?bn1?n2bn?bn1?n2bn1?2nbn1n?12???1nbn2?2nbn?bn?2?2bn?4?an?1n2?1?1?2?21n2?1 显然，对于1?n?0，有a*n?an?1?4,n?N* (2)证明：设 1ncn?2cn?1?bn?1bn,n?N，则 1?1?1n121?n?1??1?12?1?112?n?2?2?n?n?1??????n1n22?1?11?1 2?1??n?1??2n?1?n?1?2??2n?11n???21n?1??2?2?12?2n?2?1?1??12n?1??2?2?n?1?1?1?2n???2n?1??n?2??2?n?1??n?0,?cn?21n?2,n?N* *设SnSn??c1?c2???cn,n?N*，则当n?2k?2?1?k?N?时， 13122?14???12312?1k??11???2?34k11??11???1???3???k?1???k????22??2?12???2?1?2??2?2???2k?1?12k?k?12。 所以，取n0?b??1?2???b1????24009?2，对?n?n0都有： ?4017?1??Sn?Sn??20080?2???b?b?1?3?????1?n?1???b2?bn?? 故有b2b1?b3b2???bnbn?1?bn?1bn f(x)?x?bx?c(b?1,c?R)n 例23.(2007年泉州市高三质检) 已知函数 2，若f(x)的定 n义域为[－1，0]，值域也为[－1，0].若数列{bn}满足b?f(n)n3(n?N)*，记数列{b}的前n项？并证明你的结论。 和为Tn，问是否存在正常数A，使得对于任意正整数n都有T 8 n?A 解析:首先求出f(x)?∴Tn12k?122x?2x,∵b?f(n)?n?2n?1n33 141,125nnn?b1?b2?b3???bn?1?1???12k12??1213???1n,∵1?314?2???16?17?18?4?18?12,… ?1?2k?1?2?2k?1?12k,故当n?2k时,Tn?k2?1, 因此，对任何常数A，设m是不小于A的最小正整数， 则当n?22m?2时,必有T故不存在常数A使Tn?2m?22?1?m?A. n?A对所有n?2的正整数恒成立. 表示的平面区域为D,设D内整 nnx?0, 例24.(2008年中学教学参考)设不等式组???y?0,?y??nx?3n?数坐标点的个数为an.设当nSn?1an?1?1an?2???1a2n, . 1a2n7n?1136?2时,求证:1?1?1???1?7n?11a1a2a3a2n1a236 解析:容易得到aS2n?1?12n?3n,所以,要证1a1??1a3????只要证 171812n?112?13???12n?7n?1112327,因为S(n?1)?2n?1?12?(13?14)?(15?16??)???(?1?12n?1?2???12n ?1??T21?T22???T2n?1??7n?1112,所以原命题得证. 12五、迭代放缩 例25. 已知xn?1?xn?4xn?1,x1?1,求证:当n?2时, 12n?1n?|xi?1i?2|?2?21?n 解析:通过迭代的方法得到 例26. 设S 解析: ?|sin1!21xn?2?,然后相加就可以得到结论 1 k,若k≥n恒有:|Sn+k－Sn|< n n??sin2!22???sinn!,求证:对任意的正整数2n|Sn?k?Sn|?|sin(n?1)!2n?1?sin(n?2)!2n?2???sin(n?k)2n?k| 1???12n?ksin(n?1)!212nn?1|?|sin(n?2)!212kn?2|???|sinn(?k)212nn?k|?12n?1? 2n?2?(12?122???)?12n?(1?12k)? 所以|S12n 又2n?(1?1)n?Cn?Cn???Cn?n01nn?k?Sn|??1n 六、借助数列递推关系 例27.求证:1?21?32?4?1?3?52?4?6???1?3?5???(2n?1)2?4?6???2n?2n?2?1 解析: 设an?1?3?5???(2n?1)2?4?6???2n则 9 an?1?2n?12(n?1)an?2(n?1)an?1?2nan?an,从而 an?2(n?1)an?1?2nan,相加后就可以得到 12n?312n?2a1?a2???an?2(n?1)an?1?2a1?2(n?1)??1?(2n?2)??1 所以1?21?32?4?1?3?52?4?6???1?3?5???(2n?1)2?4?6???2n?2n?2?1 例28. 求证:1?21?32?4?1?3?52?4?6???1?3?5???(2n?1)2?4?6???2n?2n?1?1 解析: 设aan?1?2n?12(n?1)n?1?3?5???(2n?1)2?4?6???2n则 ,从而 an?[2(n?1)?1]an?1?(2n?1)an?an?1an?1?[2(n?1)?1]an?1?(2n?1)an,相加后就可以得到 12n?132a1?a2???an?(2n?1)an?1?3a1?(2n?1)???2n?1?1 例29. 若a1 解析: ?1,an?1?an?n?1,求证:1a1?1a2???1an?2(n?1?1) an?2?an?1?n?2?an?an?1?1?1an?1?an?2?an 所以就有 1a1?1a2???1an?1a1?an?1?an?a2?a1?2an?1an?a2?2n?1?2 七、分类讨论 例30.已知数列{a}的前n项和S满足Snnn?2an?(?1),n?1.证明：对任意的整数 nm?4，有 1a4?1a5???1am?78 ?2?., 解析:容易得到an?23n?2?(?1)n?1 由于通项中含有(?1)，很难直接放缩，考虑分项讨论： 当n?3且n为奇数时1131 132?2n?2n?1nan?an?1?22(n?2?1?2n?1?1)?22?2n?3?2n?1?2n?2?1 ?32?2?2n?322n?2n?1?32?(12n?2?12n?1)（减项放缩），于是 1a412 ①当m?4且m为偶数时 ?12?3(13??1a5???1am?1a4?(1a5?1a6)???(1am?1?1am) 22?124???12m?2)?311137??(1?m?4)???. 2428821a4?1a5???1am?1a4?1a5???1am?1am?1②当m?4且m为奇数时 1a41a51am1am?178（添项放缩）由①知 ??????.由①②得证。 10