Source: IMO ShortList 1973, Sweden 1, IMO 1973, Day 2, Problem 3
Tags: Sequence, algebra, Inequality, construction, geometric sequence, IMO, IMO 1973
Let $a_1, \ldots, a_n$ be $n$ positive numbers and $0 < q < 1.$ Determine $n$ positive numbers $b_1, \ldots, b_n$ so that:
a.) $ a_{k} < b_{k}$ for all $k = 1, \ldots, n,$
b.) $q < \frac{b_{k+1}}{b_{k}} < \frac{1}{q}$ for all $k = 1, \ldots, n-1,$
c.) $\sum \limits^n_{k=1} b_k < \frac{1+q}{1-q} \cdot \sum \limits^n_{k=1} a_k.$
LaTeX is very boring, so I´ll just send the cases $ n = 5$ and $ n = 6$.
$ b_1 = a_1 + q(a_5 + a_2) + q^2(a_4 + a_3)$
$ b_2 = a_2 + q(a_1 + a_3) + q^2(a_5 + a_4)$
$ b_3 = a_3 + q(a_2 + a_4) + q^2(a_1 + a_5)$
$ b_4 = a_4 + q(a_3 + a_5) + q^2(a_2 + a_1)$
$ b_5 = a_5 + q(a_4 + a_1) + q^2(a_3 + a_2)$
$ b_1 = a_1 + q(a_6 + a_2) + q^2(a_5 + a_3) + q^3a_4$
$ b_2 = a_2 + q(a_1 + a_3) + q^2(a_6 + a_4) + q^3a_5$
$ b_3 = a_3 + q(a_2 + a_4) + q^2(a_1 + a_5) + q^3a_6$
$ b_4 = a_4 + q(a_3 + a_5) + q^2(a_2 + a_6) + q^3a_1$
$ b_5 = a_5 + q(a_4 + a_6) + q^2(a_3 + a_1) + q^3a_2$
$ b_6 = a_6 + q(a_5 + a_1) + q^2(a_4 + a_2) + q^3a_3$
I hope there are no difficulties in generalizing. In $ b_i$, indexes wich are next to $ i$ are monitored by $ q$´s with low exponents. To see that all conditions are fulfilled, observe we have
$ \frac {1 + q}{1 - q} = 1 + 2\sum \limits^\infty_{i = 1} q^i$.