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# Minimum value

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?

Note by Wildan Bagus W. Hafidz
7 months, 2 weeks ago

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

- 6 months, 3 weeks ago