2001 USAMO Problems/Problem 3: Difference between revisions

Revision as of 17:49, 31 May 2011

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

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

Show that

$0 \le ab + bc + ca - abc \leq 2.$

First we prove the lower bound.

Note that we cannot have $a, b, c$ all greater than 1. Therefore, suppose $a \le 1$ . Then $ab + bc + ca - abc = a(b + c) + bc(1-a) \ge 0.$

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

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

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,

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

This completes the proof.

@@ Line 6: / Line 6: @@
 == Solution ==
-{{solution}}
+First we prove the lower bound.
+Note that we cannot have <math>a, b, c</math> all greater than 1.
+Therefore, suppose <math>a \le 1</math>.
+Then
+<cmath>ab + bc + ca - abc = a(b + c) + bc(1-a) \ge 0.</cmath>
 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 terms of <math>b</math> and <math>c</math>,
@@ Line 17: / Line 22: @@
 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