\(\frac { x }{ y } +\frac { y }{ z } +\frac { z }{ x } =\frac { y }{ x } +\frac { z }{ y } +\frac { x }{ z } =x+y+z=3\)

This a part of my set NMTC 2nd Level (Junior) held in 2014.

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## Comments

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TopNewest\(\dfrac{x}{y}+\dfrac{y}{z}+\dfrac{z}{x}\ge 3\) by AM-GM, with equality case if and only if \(x=y=z\)

Since \(x+y+z=3\), then we know the only real solution is \((x,y,z)=\boxed{(1,1,1)}\)

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Good observation.

Note that the application of AM-GM assumes that the variables are positive.

Are there solutions over all reals?

What about over all complex numbers?

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\(minimum\quad value\quad of\quad \frac { x }{ y } \quad +\quad \frac { y }{ z } +\frac { z }{ x } =3\quad \\ from\quad this\quad how\quad can\quad we\quad conclude\quad that\quad x=y=z????\)

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Hint:Make common denominator for\( \frac{ x}{y} + \frac{y}{z} + \frac{z}{x} - \frac{ y}{x} - \frac{ z}{y} - \frac{ x}{z} \).Factorize the numerator.

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\( \frac{ x}{y} + \frac{y}{z} + \frac{z}{x} - \frac{ y}{x} - \frac{ z}{y} - \frac{ x}{z} = \frac{z-y}x+\frac{x-z}y+\frac{y-x}z=\frac{(z-y)zy+(x-z)xz+(y-x)yx}{xyz}=\frac{(y-z)(x-z)(y-x)}{xyz}=0\)

Therefore \(x,y,z\ne0\) and (\(y=z\) or \(x=z\) or \(y=x\)).

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Great work.

This is a short step away from finishing the solution.

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\(\frac xy+\frac yx=2\), \(\frac xy=1\), \(x=y\).

Therefore, \(x=y=z\).

From \(x+y+z=3\), we have \(x=y=z=1\) as the only real solution.

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x=1,y=1,z=1

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Write a comment or ask a question... (1,1,1)

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How do you know that there are no other solutions?

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x=y=z=1 therefore the answer is 3 ,basic algebra question

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