1＜a≤b≤c,證明logaˇb十logbˇc+logcˇa≤logbˇa十logcˇb十logaˇc
【證明】
設x=logaˇb,y=logbˇc,
則原不等式變形為：x+y+1/(xy)≤1/x+1/y+xy,
上式通分整理得：(x-1)(y-1)(xy-1)/(xy)≥0,
因為x≥1,y≥1,所以上式顯然成立.
∴l(xiāng)ogaˇb十logbˇc+logcˇa≤logbˇa十logcˇb十logaˇc