A20 (15-6, Sweden) Given $n$ positive numbers $a_{1}, a_{2}, \cdots, a_{n}$ and a real number $q(0<q<1)$. Try to provide $n$ real numbers $b_{1}, b_{2}, \cdots, b_{n}$, such that
(a) For all positive integers $k(1 \leqslant k \leqslant n)$, we have $a_{k}<b_{k}$;
(b) For all positive integers $k(1 \leqslant k \leqslant n-1)$, we have
$$
q<\frac{b_{k+1}}{b_{k}}<\frac{1}{q} ;
$$
(c) $b_{1}+b_{2}+\cdots+b_{n}<\frac{1+q}{1-q}\left(a_{1}+a_{2}+\cdots+a_{n}\right)$.
For $k=1,2, \cdots, n$, let
$$
\begin{aligned}
b_{k}= & a_{1} q^{k-1}+a_{2} q^{k-2}+\cdots+a_{k-1} q \\
& +a_{k}+a_{k+1} q+\cdots+a_{n} q^{n-k} .
\end{aligned}
$$
(a) Clearly, $b_{k}>a_{k}$ for $k=1,2, \cdots, n$;
(b) For $k=1,2, \cdots, n-1$, we have
$$
\begin{array}{l}
q b_{k}-b_{k+1}=\left(q^{2}-1\right)\left[a_{k+1}+\cdots+a_{n} q^{n-k-1}\right] \\
<0 \text {; } \\
q b_{k+1}-b_{k}=\left(q^{2}-1\right)\left[a_{1} q^{k-1}+\cdots+a_{k}\right] \\
<0 \text {. } \\
\end{array}
$$

This implies
$$
q<\frac{b_{k+1}}{b_{k}}<\frac{1}{q} ;
$$
(c)
$$
\text { c) } \begin{aligned}
& b_{1}+b_{2}+\cdots+b_{n} \\
= & a_{1}+a_{2} q+\cdots+a_{n} q^{n-1}+a_{1} q+a_{2}+a_{3} q+\cdots+a_{n} q^{n-2} \\
& +a_{1} q^{2}+a_{2} q+a_{3}+\cdots+a_{n} q^{n-3}+\cdots+a_{1} q^{n-1} \\
& +a_{2} q^{n-2}+a_{3} q^{n-3}+\cdots+a_{n} \\
< & \left(a_{1}+a_{2}+\cdots+a_{n}\right)\left(1+2 q+2 q^{2}+\cdots+2 q^{n-1}\right) \\
= & \left(a_{1}+a_{2}+\cdots+a_{n}\right)\left(1+2 q \frac{1-q^{n-1}}{1-q}\right) \\
< & \left(a_{1}+a_{2}+\cdots+a_{n}\right)\left(1+\frac{2 q}{1-q}\right) \\
= & \frac{1+q}{1-q}\left(a_{1}+a_{2}+\cdots+a_{n}\right) .
\end{aligned}
$$

Thus, (c) is proved.