IMO 1973 · Problem 6

Let \(a_1,a_2,\ldots,a_n\) be \(n\) positive numbers, and let \(q\) be a given real number such that \(0<q<1\). Find \(n\) numbers \(b_1,b_2,\ldots,b_n\) for which

(a) \(a_k<b_k\) for \(k=1,2,\cdots,n\),

(b) \(q<\dfrac{b_{k+1}}{b_k}<\dfrac1q\) for \(k=1,2,\ldots,n-1\),

(c) \(b_1+b_2+\cdots+b_n<\dfrac{1+q}{1-q}(a_1+a_2+\cdots+a_n)\).

Official text of the problem. Official problem paper (PDF)

Answer

For \(n=1\), take \[ b_1=\frac{a_1}{1-q}. \] For \(n\ge2\), take \[ \boxed{b_k=\sum_{j=1}^{n}a_jq^{|k-j|}\qquad(k=1,\ldots,n).} \]

Solution

If \(n=1\), then \[ a_1<\frac{a_1}{1-q}<\frac{1+q}{1-q}a_1, \] because \(a_1>0\) and \(0<q<1\). Condition (b) has no instances, so this case is complete.

Now suppose \(n\ge2\), and define \(b_k\) as in the answer.

For each \(k\), the term with \(j=k\) equals \(a_k\), and all the other terms are positive. Hence \[ b_k=a_k+\sum_{\substack{1\le j\le n\\j\ne k}}a_jq^{|k-j|}>a_k, \] which proves (a).

For \(1\le k\le n-1\), cancellation gives \[ b_{k+1}-qb_k =(1-q^2)\sum_{j=k+1}^{n}a_jq^{j-k-1}>0 \] and \[ b_k-qb_{k+1} =(1-q^2)\sum_{j=1}^{k}a_jq^{k-j}>0. \] Both sums are nonempty, and every summand is positive. Since \(b_k>0\) and \(q>0\), these inequalities imply \[ q<\frac{b_{k+1}}{b_k}<\frac1q, \] proving (b).

Finally, for every \(j\in\{1,\ldots,n\}\), the finite geometric-sum formula yields \[ \begin{aligned} \sum_{k=1}^{n}q^{|k-j|} &=1+\sum_{r=1}^{j-1}q^r+\sum_{r=1}^{n-j}q^r\\ &=\frac{1+q-q^j-q^{n-j+1}}{1-q}\\ &<\frac{1+q}{1-q}. \end{aligned} \] Here an empty sum is understood as zero. Multiplying by \(a_j>0\) and summing over \(j\), we obtain \[ \sum_{k=1}^{n}b_k =\sum_{j=1}^{n}a_j\sum_{k=1}^{n}q^{|k-j|} <\frac{1+q}{1-q}\sum_{j=1}^{n}a_j. \] This proves (c), completing the construction.

The solution was checked against John Scholes' IMO solution archive (Kalva).