
6. Solve the system of equations for positive real numbers:

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


Solution: The given system reduces to

$$
z=x^{2} y+x y z, x=y^{z}+x y z, y=z^{2} x+x y z
$$

Hence

$$
z-x^{2} y=x-y^{2} z=y-z^{2} x
$$

If $x=y$, then $y^{2} z=z^{2} x$ and hence $x^{2} z=z^{2} x$. This implies that $z=x=y$. Similarly, $x=z$ implies that $x=z=y$. Hence if any two of $x, y, z$ are equal, then all are equal. Suppose no two of $x, y, z$ are equal. We may take $x$ is the largest among $x, y, z$ so that $x>y$ and $x>z$. Here we have two possibilities: $y>z$ and $z>y$.

Suppose $x>y>z$. Now $z-x^{2} y=x-y^{2} z=y-z^{2} x$ shows that

$$
y^{2} z>z^{2} x>x^{2} y
$$

But $y^{2} z>z^{2} x$ and $z^{2} x>x^{2} y$ give $y^{2}>z x$ and $z^{2}>x y$. Hence

$\left(y^{2}\right)\left(z^{2}\right)>(z x)(x y)$.

This gives $y z>x^{2}$. Thus $x^{3}y$.

Similarly, we arrive at a contradiction if $x>z>y$. The only possibility is $x=y=z$. For $x=y=z$, we get only one equation $x^{2}=1 / 2$. Since $x>0, x=1 / \sqrt{2}=y=z$. $\qquad$

