Difference between revisions of "2001 USAMO Problems/Problem 3"
Line 8: | Line 8: | ||
{{solution}} | {{solution}} | ||
− | Without | + | Without loss of generality, we assume <math>(b-1)(c-1)\ge 0</math>. From the given equation, we can express <math>a</math> in the form <math>b</math> and <math>c</math> as, |
<center> <math>a=\frac{\sqrt{(4-b^2)(4-c^2)}-bc}{2} </math></center> | <center> <math>a=\frac{\sqrt{(4-b^2)(4-c^2)}-bc}{2} </math></center> | ||
Thus, | Thus, | ||
Line 14: | Line 14: | ||
From Cauchy, | From Cauchy, | ||
− | <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)} | + | <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> |
This completes the proof. | This completes the proof. |
Revision as of 22:52, 8 February 2011
Problem
Let and satisfy
![$a^2 + b^2 + c^2 + abc = 4.$](http://latex.artofproblemsolving.com/1/6/e/16e70ab813b2e9287a1015d7b890d16f94a7073e.png)
Show that
![$ab + bc + ca - abc \leq 2.$](http://latex.artofproblemsolving.com/6/6/c/66c37e5bd5601bffeb016667562ca756e7dd1d9b.png)
Solution
This problem needs a solution. If you have a solution for it, please help us out by adding it.
Without loss of generality, we assume . From the given equation, we can express
in the form
and
as,
![$a=\frac{\sqrt{(4-b^2)(4-c^2)}-bc}{2}$](http://latex.artofproblemsolving.com/4/3/4/4344f8b39fdd47d62fae63057b6575aef0c9e7a6.png)
Thus,
![$ab + bc + ca - abc = -a (b-1)(c-1)+a+bc \le a+bc = \frac{\sqrt{(4-b^2)(4-c^2)} + bc}{2}$](http://latex.artofproblemsolving.com/c/8/d/c8d8a39e9208c03f8abfe37a83ee86e4029c3c23.png)
From Cauchy,
![$\frac{\sqrt{(4-b^2)(4-c^2)} + bc}{2} \le \frac{\sqrt{(4-b^2+b^2)(4-c^2+c^2)} }{2} = 2$](http://latex.artofproblemsolving.com/8/d/d/8dd62a83d2fd1681f136aea8227930295135e40e.png)
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 |