2001 USAMO Problems/Problem 3
Problem
Let and satisfy
![$a^2 + b^2 + c^2 + abc = 4.$](http://latex.artofproblemsolving.com/1/6/e/16e70ab813b2e9287a1015d7b890d16f94a7073e.png)
Show that
![$0 \le ab + bc + ca - abc \leq 2.$](http://latex.artofproblemsolving.com/b/f/7/bf76c4b7c8d7148354b55dc865be48fdeb231c1a.png)
Solution
First we prove the lower bound.
Note that we cannot have all greater than 1.
Therefore, suppose
.
Then
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
![$a=\frac{\sqrt{(4-b^2)(4-c^2)}-bc}{2}$](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 |