Difference between revisions of "2001 USAMO Problems/Problem 3"

Line 12: Line 12:
 
Thus,
 
Thus,
 
<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,
 +
<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)} + bc}{2} = 2</math> </center>
 +
 +
This completes the proof.
 +
  
 
== See also ==
 
== See also ==

Revision as of 22:51, 8 February 2011

Problem

Let $a, b, c \geq 0$ and satisfy

$a^2 + b^2 + c^2 + abc = 4.$

Show that

$ab + bc + ca - abc \leq 2.$

Solution

This problem needs a solution. If you have a solution for it, please help us out by adding it.

Without loosing generality, we assume $(b-1)(c-1)\ge 0$. From the given equation, we can express $a$ in the form $b$ and $c$ as,

$a=\frac{\sqrt{(4-b^2)(4-c^2)}-bc}{2}$

Thus,

$ab + bc + ca - abc = -a (b-1)(c-1)+a+bc \le a+bc = \frac{\sqrt{(4-b^2)(4-c^2)} + bc}{2}$

From Cauchy,

$\frac{\sqrt{(4-b^2)(4-c^2)} + bc}{2} \le \frac{\sqrt{(4-b^2+b^2)(4-c^2+c^2)} + bc}{2} = 2$

This completes the proof.


See also

2001 USAMO (ProblemsResources)
Preceded by
Problem 2
Followed by
Problem 4
1 2 3 4 5 6
All USAMO Problems and Solutions