Difference between revisions of "2000 IMO Problems/Problem 2"
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Let <math>a, b, c</math> be positive real numbers with <math>abc=1</math>. Show that | Let <math>a, b, c</math> be positive real numbers with <math>abc=1</math>. Show that | ||
| − | <cmath>\left( a- | + | <cmath>\left( a-1+\frac{1}{b} \right)\left( b-1+\frac{1}{c} \right)\left( c-1+\frac{1}{a} \right) \le 1</cmath> |
==Solution== | ==Solution== | ||
| − | {{ | + | There exist positive reals <math>x</math>, <math>y</math>, <math>z</math> such that <math>a = \frac{x}{y}</math>, <math>b = \frac{y}{z}</math>, <math>c = \frac{z}{x}</math>. The inequality then rewrites as <cmath>\left(\frac{x-y+z}{y}\right)\left(\frac{y-z+x}{z}\right)\left(\frac{z-x+y}{x}\right)\leq 1</cmath> |
| + | or <cmath>(x-y+z)(y-z+x)(z-x+y)\leq xyz.</cmath> Set <math>p=x-y+z</math>,<math>q=y-z+x</math>,<math>r=z-x+y</math>, we get <cmath>8pqr\leq(p+q)(q+r)(r+p).</cmath> | ||
| + | Since at most one of <math>p,q,r</math> can be negative (if 2 or more are negative, then one of <math>a,b,c</math> will become negative), for all positive we apply AM-GM, for one negative we have <math>LHS<0<RHS</math>. | ||
==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 21:26, 6 March 2024
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 
.
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 | ||
