Difference between revisions of "1973 IMO Problems/Problem 6"

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{{IMO box|year=1973|num-b=5|after=Last Question}}
 
{{IMO box|year=1973|num-b=5|after=Last Question}}
 
[[Category:Olympiad Algebra Problems]]
 
[[Category:Olympiad Algebra Problems]]
[[Category:Inequality Problems]]
 

Revision as of 17:56, 18 September 2025

Problem

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

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

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

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


Solution

We notice that the constraints are linear, in the sense that if bi is a solution for ai, q, and bi' is a solution for ai', q, then for any k, k' > 0 a solution for kai + k'ai', q is kbi + k'bi'. Also a "near" solution for ah = 1, other ai = 0 is b1 = qh-1, b2 = qh-2, ... , bh-1 = q, bh = 1, bh+1 = q, ... , bn = qn-h. "Near" because the inequalities in (a) and (b) are not strict.

However, we might reasonably hope that the inequalities would become strict in the linear combination, and indeed that is true. Define br = qr-1a1 + qr-2a2 + ... + qar-1 + ar + qar+1 + ... + qn-ran. Then we may easily verify that (a) - (c) hold.


See Also

1973 IMO (Problems) • Resources
Preceded by
Problem 5
1 2 3 4 5 6 Followed by
Last Question
All IMO Problems and Solutions