2001 USAMO Problems/Problem 3

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Problem

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

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

Show that

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

Solution

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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,

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

See also

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