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Problem A. 391. (January 2006)

A. 391. Construct a sequence a1,a2,...,aN of positive reals such that n1an0+n2an1+...+nkank-1>2.7(a1+a2+...+aN) for arbitrary integers 1=n0<n1<...<nk=N.

(5 pont)

Deadline expired on February 15, 2006.


Solution. Let N be sufficiently large (it will be defined later) and set a_n=\frac1n. Then

a_1+a_2+\dots+a_N=1+\frac12+\frac13+\dots+\frac1N>1+\log N

and for any sequence 1=n0<n1<...<nk=N of indices we have


n_1a_{n_0}+n_2a_{n_1}+\ldots+n_ka_{n_{k-1}}=
\frac{n_1}{n_0}+\frac{n_2}{n_1}+\dots+\frac{n_k}{n_{k-1}} \ge

\ge
k\root{k}\of{\frac{n_1}{n_0}\cdot\frac{n_2}{n_1}\cdot\dots\cdot\frac{n_k}{n_{k-1}}}
=k\root{k}\of{\frac{n_k}{n_0}}= k\root{k}\of{N}=\frac{N^{1/k}}{1/k}.

The minimum of the function \frac{N^t}{t} is at point t=\frac1{\log N}, its value is e.log N. Therefore,


n_1a_{n_0}+n_2a_{n_1}+\ldots+n_ka_{n_{k-1}}\ge
\frac{N^{1/k}}{1/k}\ge e\cdot \log N.

If N is chosen such that \log N>\frac{2,7}{e-2,7} then e.log N>2,7.(1+log N).


Statistics:

4 students sent a solution.
5 points:Paulin Roland, Tomon István.
0 point:2 students.

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