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# Problem 5b

Solve for $$x,y,z$$
$$\frac { x }{ y } +\frac { y }{ z } +\frac { z }{ x } =\frac { y }{ x } +\frac { z }{ y } +\frac { x }{ z } =x+y+z=3$$
This a part of my set NMTC 2nd Level (Junior) held in 2014.

Note by Siddharth G
2 years, 1 month ago

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$$\dfrac{x}{y}+\dfrac{y}{z}+\dfrac{z}{x}\ge 3$$ 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 $$(x,y,z)=\boxed{(1,1,1)}$$ · 2 years, 1 month ago

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? Staff · 2 years, 1 month ago

$$minimum\quad value\quad of\quad \frac { x }{ y } \quad +\quad \frac { y }{ z } +\frac { z }{ x } =3\quad \\ from\quad this\quad how\quad can\quad we\quad conclude\quad that\quad x=y=z????$$ · 2 years ago

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. Staff · 2 years, 1 month ago

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

Therefore $$x,y,z\ne0$$ and ($$y=z$$ or $$x=z$$ or $$y=x$$). · 2 years, 1 month ago

Great work.

This is a short step away from finishing the solution. Staff · 2 years, 1 month ago

When $$y=z$$, $$\frac xy+1+\frac yx=\frac yx+1+\frac xy=x+2y=3$$

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

Therefore, $$x=y=z$$.

From $$x+y+z=3$$, we have $$x=y=z=1$$ as the only real solution. · 2 years ago

x=1,y=1,z=1 · 2 years, 1 month ago

x=y=z=1 therefore the answer is 3 ,basic algebra question · 2 years, 1 month ago

Write a comment or ask a question... (1,1,1) · 2 years, 1 month ago