
4. Find all positive real numbers $x, y, z$ such that

$$
2 x-2 y+\frac{1}{z}=\frac{1}{2014}, \quad 2 y-2 z+\frac{1}{x}=\frac{1}{2014}, \quad 2 z-2 x+\frac{1}{y}=\frac{1}{2014}
$$


Solution: Adding the three equations, we get

$$
\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{3}{2014}
$$

We can also write the equations in the form

$$
2 z x-2 z y+1=\frac{z}{2014}, \quad 2 x y-2 x z+1=\frac{x}{2014}, \quad 2 y z-2 y x+1=\frac{y}{2014}
$$

Adding these, we also get

$$
2014 \times 3=x+y+z
$$

Therefore

$$
\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)(x+y+z)=\frac{3}{2014} \times(2014 \times 3)=9
$$

Using AM-GM inequality, we therefore obtain

$$
9=\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)(x+y+z) \geq 9 \times(x y z)^{1 / 3}\left(\frac{1}{x y z}\right)^{1 / 3}=9
$$

Hence equality holds in AM-GM inequality and we conclude $x=y=z$. Thus

$$
\frac{1}{x}=\frac{1}{2014}
$$

which gives $x=2014$. We conclude

$$
x=2014, \quad y=2014, \quad z=2014
$$
