2001 USAMO Problems/Problem 3: Difference between revisions

Revision as of 17:46, 31 May 2011

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

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

Show that

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

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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 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 3: / Line 3: @@
 <center><math>a^2 + b^2 + c^2 + abc = 4.</math></center>
 Show that
-<center><math>ab + bc + ca - abc \leq 2.</math></center>
+<center><math>0 \le ab + bc + ca - abc \leq 2.</math></center>
 == Solution ==

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