Difference between revisions of "2000 IMO Problems/Problem 2"
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==Video Solution== | ==Video Solution== | ||
https://youtu.be/-JyUrRq18BU?si=_kVNesHwbqI4_P9d [little-fermat] | https://youtu.be/-JyUrRq18BU?si=_kVNesHwbqI4_P9d [little-fermat] | ||
| + | |||
| + | ==Video Solution 2== | ||
| + | |||
| + | By Brewsly, simple and elegant: https://youtu.be/OpPkgOs38ao | ||
==See Also== | ==See Also== | ||
{{IMO box|year=2000|num-b=1|num-a=3}} | {{IMO box|year=2000|num-b=1|num-a=3}} | ||
Latest revision as of 21:56, 19 August 2025
Problem
Let 
be positive real numbers with 
. Show that
![\[\left( a-1+\frac{1}{b} \right)\left( b-1+\frac{1}{c} \right)\left( c-1+\frac{1}{a} \right) \le 1\]](http://latex.artofproblemsolving.com/e/9/6/e96e8028ff8719b9a661de10c0cb70fc7b43ca67.png)
Solution
There exist positive reals 
, 
, 
such that 
, 
, 
. The inequality then rewrites as ![\[\left(\frac{x-y+z}{y}\right)\left(\frac{y-z+x}{z}\right)\left(\frac{z-x+y}{x}\right)\leq 1\]](http://latex.artofproblemsolving.com/6/5/d/65d590ef6465d23c4bb734a0d8868c815358315c.png)
or ![\[(x-y+z)(y-z+x)(z-x+y)\leq xyz.\]](http://latex.artofproblemsolving.com/a/3/b/a3b666d38d4e1e88ba32217baccc04455acf9398.png)
Set 
,
,
, we get ![\[8pqr\leq(p+q)(q+r)(r+p).\]](http://latex.artofproblemsolving.com/3/9/7/3978bbd0d246e2d945de2d0a5608e804255ca2d2.png)
Since at most one of 
can be negative (if 2 or more are negative, then one of 
will become negative), for all positive we apply AM-GM, for one negative we have 
.
Video Solution
https://youtu.be/-JyUrRq18BU?si=_kVNesHwbqI4_P9d [little-fermat]
Video Solution 2
By Brewsly, simple and elegant: https://youtu.be/OpPkgOs38ao
See Also
| 2000 IMO (Problems) • Resources | ||
| Preceded by Problem 1 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 3 |
| All IMO Problems and Solutions | ||
