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