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

Revision as of 21:51, 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,

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

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.

@@ Line 12: / Line 12: @@
 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>
+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 ==

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