Difference between revisions of "1973 IMO Problems/Problem 6"
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==Problem== | ==Problem== | ||
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Let <math>a_1, a_2,\cdots, a_n</math> be <math>n</math> positive numbers, and let <math>q</math> be a given real number such that <math>0<q<1.</math> Find <math>n</math> numbers <math>b_1, b_2, \cdots, b_n</math> for which | Let <math>a_1, a_2,\cdots, a_n</math> be <math>n</math> positive numbers, and let <math>q</math> be a given real number such that <math>0<q<1.</math> Find <math>n</math> numbers <math>b_1, b_2, \cdots, b_n</math> for which | ||
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(c) <math>b_1+b_2+\cdots+b_n<\dfrac{1+q}{1-q}(a_1+a_2+\cdots+a_n).</math> | (c) <math>b_1+b_2+\cdots+b_n<\dfrac{1+q}{1-q}(a_1+a_2+\cdots+a_n).</math> | ||
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==Solution== | ==Solution== | ||
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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. | 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. | ||
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==See Also== | ==See Also== | ||
{{IMO box|year=1973|num-b=5|after=Last Question}} | {{IMO box|year=1973|num-b=5|after=Last Question}} | ||
| − | [[Category:Olympiad | + | [[Category:Olympiad Agebra Problems]] |
| + | [[Category:Inequalities Problems]] | ||
Revision as of 17:53, 18 September 2025
Problem
Let 
be 
positive numbers, and let 
be a given real number such that 
Find numbers 
for which
(a) 
for 
(b) 
for 
(c) 
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 | ||
