Given \(\displaystyle{1}{ < }{a}{ < }{b}{

limes14514qqmn

limes14514qqmn

Answered question

2022-03-24

Given 1<a<b<c prove
logalogab+logblogbc+logclogca>0.

Answer & Explanation

Carter Lin

Carter Lin

Beginner2022-03-25Added 13 answers

loga{loga{b}}+logb{logb{c}}+logc{logc{a}}>0
ln{ln{b}}ln{ln{a}}ln{a}+ln{ln{c}}ln{ln{b}}ln{b}+ln{ln{a}}ln{ln{c}}ln{c}>0
ln{ln{b}}ln{a}+ln{ln{c}}ln{b}+ln{ln{a}}ln{c}>ln{ln{a}}ln{a}+ln{ln{b}}ln{b}+ln{ln{c}}ln{c}
The last inequality is true by rearrangement inequality since ln{ln{c}}>ln{ln{b}}>ln{ln{a}} and 1ln{a}>1ln{b}>1ln{c} and the inequality is strict since the numbers are distinct.
Wilson Rivas

Wilson Rivas

Beginner2022-03-26Added 12 answers

Let us use the property of log.
We know that logab=log10blog10a
If we apply that in your problem, we end up with
log10b(log10a)2+log10c(log10b)2+log10a(log10c)2
Now we can able to see that the quantity is >0

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