Problem 5b

Solve for x,y,zx,y,z
xy+yz+zx=yx+zy+xz=x+y+z=3\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
4 years, 10 months ago

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xy+yz+zx3\dfrac{x}{y}+\dfrac{y}{z}+\dfrac{z}{x}\ge 3 by AM-GM, with equality case if and only if x=y=zx=y=z

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

Daniel Liu - 4 years, 10 months ago

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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?

Calvin Lin Staff - 4 years, 10 months ago

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minimumvalueofxy+yz+zx=3fromthishowcanweconcludethatx=y=z????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????

Prashant Ramnani - 4 years, 9 months ago

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Hint: Make common denominator forxy+yz+zxyxzyxz \frac{ x}{y} + \frac{y}{z} + \frac{z}{x} - \frac{ y}{x} - \frac{ z}{y} - \frac{ x}{z} .
Factorize the numerator.

Calvin Lin Staff - 4 years, 10 months ago

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xy+yz+zxyxzyxz=zyx+xzy+yxz=(zy)zy+(xz)xz+(yx)yxxyz=(yz)(xz)(yx)xyz=0 \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,z0x,y,z\ne0 and (y=zy=z or x=zx=z or y=xy=x).

Kenny Lau - 4 years, 10 months ago

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Great work.

This is a short step away from finishing the solution.

Calvin Lin Staff - 4 years, 10 months ago

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@Calvin Lin When y=zy=z, xy+1+yx=yx+1+xy=x+2y=3\frac xy+1+\frac yx=\frac yx+1+\frac xy=x+2y=3

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

Therefore, x=y=zx=y=z.

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

Kenny Lau - 4 years, 9 months ago

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x=1,y=1,z=1

deepak bisht - 4 years, 10 months ago

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

John Wilcox - 4 years, 10 months ago

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How do you know that there are no other solutions?

Calvin Lin Staff - 4 years, 10 months ago

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

Praguna Manvi - 4 years, 10 months ago

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