2001 USAMO Problems/Problem 3: Difference between revisions

Revision as of 21:46, 8 February 2011

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

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

Show that

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

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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}

@@ Line 8: / Line 8: @@
 {{solution}}
-Without loosing 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>,
+Without loosing 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>
+Thus,
+<center> $ab + bc + ca - abc = -a (b-1)(c-1)+a+bc \le a+bc = \frac{\sqrt{(4-b^2)(4-c^2)} + bc}{2} </center>
 == 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