Difference between revisions of "1995 IMO Problems/Problem 4"
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− | First we start by solving for <math>x_{i}</math> in the | + | First we start by solving for <math>x_{i}</math> in the recursive relation |
<math>x_{i-1}+\frac{2}{x_{i-1}}=2x_{i}+\frac{1}{x_{i}}</math> | <math>x_{i-1}+\frac{2}{x_{i-1}}=2x_{i}+\frac{1}{x_{i}}</math> | ||
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~ Tomas Diaz. orders@tomasdiaz.com | ~ Tomas Diaz. orders@tomasdiaz.com | ||
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{{alternate solutions}} | {{alternate solutions}} |
Revision as of 19:30, 18 November 2023
Problem
The positive real numbers satisfy the relations
and
for
Find the maximum value that can have.
Solution
First we start by solving for in the recursive relation
or
So we have two recursive properties to chose from.
If we want to maximize then we can use from to . This will make the largest and the smallest.
Then we can simply use to get since the reciprocal will make it very large.
Then we use and solve for
This means that we can write as:
Then ,
Solving for we get:
Since all are defined as pisitive then,
~ Tomas Diaz. orders@tomasdiaz.com
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
See Also
1995 IMO (Problems) • Resources | ||
Preceded by Problem 3 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 5 |
All IMO Problems and Solutions |