Tuesday, 25 November 2014

inequality - Prove ln(n)ltn using only log(xy)=ylog(x)




Prove ln(n)<n(nN)
Using only the rule log(xy)=ylog(x)
I tried using any of the known Inequalities am-gm, power mean, titu's lemma, holder's,... but I seem I can't go anywhere near required.
Any hint?



Edit
It was simpler than I thought, it's done with Bernoulli's Inequality.
en>2n=(1+1)n1+n>n
ln(en)>ln(n)
ln(n)<n


Answer




Your inequality could be generalized as lnxx1,x>0 To prove it, you perhaps have to use calculus, not just elementary mathematics.Notice that, for f(x)=lnxx+1,(x>0), f(x)=1x1. Thus, f(x)>0, when $01.Forthereason,f(x)reachesitsmaximumvaluef(1)=0atx=0,namely,f(x) \leq 0$.



But if you have known Bernoulli's inequality, there indeed exists an elementary proof.



Since




(1+x)n1+nx,(x1,nN+).





Let x=e1. Then en1+n(e1)=1+nen>n.



It follows that n>lnn.


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