Difference between revisions of "2000 JBMO Problems/Problem 1"
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== Solution 2 == | == Solution 2 == |
Latest revision as of 16:17, 24 March 2024
Contents
Problem
Let and
be positive reals such that
Show that
.
Solution 1
After some manipulation we get:
Case 1:
, So
has to be
, so
or
Case 2:
, So
has to be
, so
or
Thus
(This is wrong :|)
Solution 2
Rearranging the equation yields
If
in the large equation, then
must be a factor of the large equation. Note that we can rewrite the large equation as
We can factor the difference of cubes in the first part and factor
in the second part, resulting in
Finally, we can factor by grouping, which results in
By the Zero Product Property, either
or
However, since
and
are both positive,
can not equal zero, so we have proved that
See Also
2000 JBMO (Problems • Resources) | ||
Preceded by First Problem |
Followed by Problem 2 | |
1 • 2 • 3 • 4 | ||
All JBMO Problems and Solutions |