Xupeisen110高中数学1不等式证明综合练****教材:不等式证明综合练****目的:系统小结不等式证明的几种常用方法,渗透“化归”“类比”“换元”等数学思想。过程:一、简述不等式证明的几种常用方法比较、综合、分析、换元、反证、放缩、构造二、例一、已知0<x<1,0<a<1,试比较|)1(log||)1(log|xxaa??和的大小。解一:????)1(log)1(log)1(log)1(log|)1(log||)1(log|22xxxxxxaaaaaa??????????xxxaa????11log)1(log2∵0<1?x2<1,1110????xx∴011log)1(log2????xxxaa∴|)1(log||)1(log|xxaa???解二:2111111log11log)1(log)1(log)1(log)1(logxxxxxxxxxxxaa????????????????)1(log121xx????∵0<1?x2<1,1+x>1,∴0)1(log21????xx∴1)1(log121????xx∴|)1(log||)1(log|xxaa???解三:∵0<x<1,∴0<1?x<1,1<1+x<2,∴0)1(log,0)1(log????xxaa∴左?右=)1(log)1(log)1(log2xxxaaa?????∵0<1?x2<1,且0<a<1∴0)1(log2??xa∴|)1(log||)1(log|xxaa???Xupeisen110高中数学2变题:若将a的取值范围改为a>0且a?1,其余条件不变。例二、已知x2=a2+b2,y2=c2+d2,且所有字母均为正,求证:xy≥ac+bd证一:(分析法)∵a,b,c,d,x,y都是正数∴要证:xy≥ac+bd只需证:(xy)2≥(ac+bd)2即:(a2+b2)(c2+d2)≥a2c2+b2d2+2abcd展开得:a2c2+b2d2+a2d2+b2c2≥a2c2+b2d2+2abcd即:a2d2+b2c2≥2abcd由基本不等式,显然成立∴xy≥ac+bd证二:(综合法)xy=222222222222dbdacbcadcba??????≥bdacbdacdbabcdca??????22222)(2证三:(三角代换法)∵x2=a2+b2,∴不妨设a=xsin?,b=xcos?y2=c2+d2c=ysin?,d=ycos?∴ac+bd=xysin?sin?+xycos?cos?=xycos(???)≤xy例三、已知x1,x2均为正数,求证:2212
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