设1<a≤b≤c,证明logaˇb十logbˇc≤logcˇa≤logbˇa十logcˇb十logaˇc

来源:学生作业帮助网 编辑:作业帮 时间:2024/08/10 00:36:50
设1<a≤b≤c,证明logaˇb十logbˇc≤logcˇa≤logbˇa十logcˇb十logaˇc
x){n=su.IdOO<ݞJ:ݞ O'BX@Dl2\PGMR> /[2Qx7La",0*l&TBM=ov>ӽ~Ϭ JmC}JM@NvE%Dߓ]@/f=h{6u mOMjҨ5Ԩ Ժ W u*Ά'B{6cm:&<_ݭVc A_\g8Sr@

设1<a≤b≤c,证明logaˇb十logbˇc≤logcˇa≤logbˇa十logcˇb十logaˇc
设1<a≤b≤c,证明logaˇb十logbˇc≤logcˇa≤logbˇa十logcˇb十logaˇc

设1<a≤b≤c,证明logaˇb十logbˇc≤logcˇa≤logbˇa十logcˇb十logaˇc
1<a≤b≤c,证明logaˇb十logbˇc+logcˇa≤logbˇa十logcˇb十logaˇc
【证明】
设x=logaˇb,y=logbˇc,
则原不等式变形为:x+y+1/(xy)≤1/x+1/y+xy,
上式通分整理得:(x-1)(y-1)(xy-1)/(xy)≥0,
因为x≥1,y≥1,所以上式显然成立.
∴logaˇb十logbˇc+logcˇa≤logbˇa十logcˇb十logaˇc