Define

$$
f(x, y, z)=\frac{(x y+y z+z x)(x+y+z)}{(x+y)(x+z)(y+z)} .
$$

Determine the set of real numbers \( r \) for which there exists a triplet \((x, y, z)\) of positive real numbers satisfying \( f(x, y, z)=r \).
We prove that $1<f(x, y, z) \leq \frac{9}{8}$, and that $f(x, y, z)$ can take on any value within the range $\left(1, \frac{9}{8}\right]$.

The expression for $f(x, y, z)$ can be simplified to

$$
f(x, y, z)=1+\frac{x y z}{(x+y)(x+z)(y+z)} .
$$

Since $x, y, z$ are positive, we get $1<f(x, y, z)$.

The inequality $f(x, y, z) \leq \frac{9}{8}$ can be simplified to

$$
x^{2} y+x^{2} z+y^{2} x+y^{2} z+z^{2} x+z^{2} y-6 x y z \geq 0 .
$$

Rearrange the left hand side as follows:

$$
\begin{aligned}
& x^{2} y+x^{2} z+y^{2} x+y^{2} z+z^{2} x+z^{2} y-6 x y z= \\
& x\left(y^{2}+z^{2}\right)-2 x y z+y\left(x^{2}+z^{2}\right)-2 x y z+z\left(x^{2}+y^{2}\right)-2 x y z= \\
& x(y-z)^{2}+y(x-z)^{2}+z(x-y)^{2} .
\end{aligned}
$$

This expression is clearly non-negative when $x, y, z$ are non-negative. To prove that $f(x, y, z)$ takes any values in the interval $\left(1, \frac{9}{8}\right]$, define

$$
g(t)=f(t, 1,1)=1+\frac{t}{2(1+t)^{2}}
$$

Then $g(1)=\frac{9}{8}$ and $g(t)$ approaches 1 as $t$ approaches 0. It follows from the continuity of $g(t)$ for $0<t \leq 1$ that it takes all values in the interval $\left(1, \frac{9}{8}\right]$. (Alternatively, one can check that the quadratic equation $g(t)=r$ has a solution $t$ for any number $r$ in the interval $\left(1, \frac{9}{8}\right]$.)