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

Revision as of 12:03, 17 September 2012

Problem

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

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

Show that

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

Solution

First we prove the lower bound.

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

Now, without loss of generality, we assume that $b$ and $c$ are either both greater than 1 or both less than one, so $(b-1)(c-1)\ge 0$ . From the given equation, we can express $a$ in terms of $b$ 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 the Cauchy-Schwarz Inequality,

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

This completes the proof.

@@ Line 26: / Line 26: @@
 {{USAMO newbox|year=2001|num-b=2|num-a=4}}
+[[Category:Olympiad Algebra Problems]]
 [[Category:Olympiad Inequality Problems]]

2001 USAMO (Problems • Resources)
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Difference between revisions of "2001 USAMO Problems/Problem 3"

Revision as of 12:03, 17 September 2012

Problem

Solution

See also