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

Revision as of 21:52, 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 loss of 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,

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

This completes the proof.

@@ 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> as,
+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>
 Thus,
@@ Line 14: / Line 14: @@
 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>
+<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.

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