Difference between revisions of "2001 USAMO Problems/Problem 3"
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<center> <math>ab + bc + ca - abc = -a (b-1)(c-1)+a+bc \le a+bc = \frac{\sqrt{(4-b^2)(4-c^2)} + bc}{2}</math> </center> | <center> <math>ab + bc + ca - abc = -a (b-1)(c-1)+a+bc \le a+bc = \frac{\sqrt{(4-b^2)(4-c^2)} + bc}{2}</math> </center> | ||
| − | From Cauchy, | + | From the [[Cauchy-Schwarz Inequality]], |
<center> <math> \frac{\sqrt{(4-b^2)(4-c^2)} + bc}{2} \le \frac{\sqrt{(4-b^2+b^2)(4-c^2+c^2)} }{2} = 2</math> </center> | <center> <math> \frac{\sqrt{(4-b^2)(4-c^2)} + bc}{2} \le \frac{\sqrt{(4-b^2+b^2)(4-c^2+c^2)} }{2} = 2</math> </center> | ||
Revision as of 17:52, 31 May 2011
Problem
Let 
and satisfy
Show that
Solution
First we prove the lower bound.
Note that we cannot have 
all greater than 1.
Therefore, suppose 
.
Then
![\[ab + bc + ca - abc = a(b + c) + bc(1-a) \ge 0.\]](http://latex.artofproblemsolving.com/f/0/d/f0dbb77c9c53e5e7312d70d0a66a297305c45e44.png)
Now, without loss of generality, we assume that 
and 
are either both greater than 1 or both less than one, so 
. From the given equation, we can express 
in terms of and as
Thus,
From the Cauchy-Schwarz Inequality,
This completes the proof.
See also
| 2001 USAMO (Problems • Resources) | ||
| Preceded by Problem 2 |
Followed by Problem 4 | |
| 1 • 2 • 3 • 4 • 5 • 6 | ||
| All USAMO Problems and Solutions | ||
