Difference between revisions of "2016 OIM Problems/Problem 2"
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== Solution == | == Solution == | ||
− | {{ | + | Clearly <math>\boxed{(1,1,1)}</math> works. We then prove that no other solutions exist. |
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+ | <math>(1)</math> First, if <math>x>1</math>, then <math>y^2+y-1=\frac{1}{x}<1</math>, so <math>y^2+y-2<0</math> and thus <math>y\in(-2,1)</math>. Since <math>y</math> must be positive, then we have <math>0<y<1</math>. | ||
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+ | <math>(2)</math> Next, if <math>0<x<1</math>, then <math>y^2+y-1=\frac{1}{x}>1</math>, so <math>y^2+y-2>0</math> and thus <math>y\in(-\infty,-2)\cup(1,\infty)</math>. Since <math>y</math> must be positive, then we have <math>y>1</math>. | ||
+ | |||
+ | Since <math>x=1</math> implies that <math>y=1</math> and <math>z=1</math> because they are all positive, this covers all possible values of <math>x</math>. Now, we use these properties on all variables. | ||
+ | |||
+ | Assume that <math>0<x<1</math>. By <math>(2)</math>, we have <math>y>1</math>, and by <math>(1)</math>, we have <math>0<z<1</math>. Finally, by <math>(2)</math>, we have <math>x>1</math>, a contradiction, so no solutions exist in this case. | ||
+ | |||
+ | Similarly, assume that <math>x>1</math>. By <math>(1)</math>, we have <math>0<y<1</math>, and by <math>(2)</math>, we have <math>z>1</math>. Finally, by <math>(1)</math>, we have <math>0<x<1</math>, a contradiction, so no solutions exist in this case either. | ||
+ | |||
+ | As a result, this leaves <math>\boxed{(1,1,1)}</math> as our only solution. | ||
== See also == | == See also == | ||
[[OIM Problems and Solutions]] | [[OIM Problems and Solutions]] |
Revision as of 17:20, 23 March 2025
Problem
Find all positive real solutions of the system of equations:
~translated into English by Tomas Diaz. ~orders@tomasdiaz.com
Solution
Clearly works. We then prove that no other solutions exist.
First, if
, then
, so
and thus
. Since
must be positive, then we have
.
Next, if
, then
, so
and thus
. Since
must be positive, then we have
.
Since implies that
and
because they are all positive, this covers all possible values of
. Now, we use these properties on all variables.
Assume that . By
, we have
, and by
, we have
. Finally, by
, we have
, a contradiction, so no solutions exist in this case.
Similarly, assume that . By
, we have
, and by
, we have
. Finally, by
, we have
, a contradiction, so no solutions exist in this case either.
As a result, this leaves as our only solution.