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

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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lo

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