Given \(x\), \(y\), and \(z\) are real numbers that satisfy \(x + y + z = \frac { 1 }{ x } +\frac { 1 }{ y } +\frac { 1 }{ z } \) and \(xyz = 1\). Find the smallest value of \(|x + y + z|\).

*My Way*

Based on \(QM \ge AM\),

\(\frac { 3 }{ \frac { 1 }{ x } +\frac { 1 }{ y } +\frac { 1 }{ z } } \ge \frac { x+y+z }{ 3 } \\ 9\ge \left( \frac { 1 }{ x } +\frac { 1 }{ y } +\frac { 1 }{ z } \right) \left( x+y+z \right) \)

Because \(x + y + z = \frac { 1 }{ x } +\frac { 1 }{ y } +\frac { 1 }{ z } \),

\(9\ge \left( \frac { 1 }{ x } +\frac { 1 }{ y } +\frac { 1 }{ z } \right) \left( x+y+z \right) \\ 9\ge { \left( x+y+z \right) }^{ 2 }\\ \pm 3\ge x+y+z\\ 3\ge |x+y+z|\)

But, I found the maximum value. Can you give me a hint?

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## Comments

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TopNewestSeems like 1 is the minimum value. However, I'm stuck atm.

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