Solution:
$$f(x, y, z) = (x + y + z)(-x + y + z)(x - y + z)(x + y - z) \quad \left[= x^{2} y^{2} + y^{2} z^{2} + z^{2} x^{2} - x^{4} - y^{4} - z^{4}\right]$$
It is easily seen that the transformation $x \mapsto -x$, and symmetrically $y \mapsto -y$ and $z \mapsto -z$, do not change $f$, so it is enough to prove the following statement: If $x, y$, and $z$ are nonnegative reals, then $x, y$, and $z$ are the sides of a triangle if and only if $f(x, y, z) > 0$.

Moreover, changing the order of $x, y$, and $z$ does not change $f$, so we may assume that $x \leq y \leq z$. Now three of the factors of $f(x, y, z)$, namely $x + y + z$, $-x + y + z$, and $x - y + z$, are clearly nonnegative.

If $x, y$, and $z$ are the sides of a triangle, the familiar triangle inequality $x + y \geq z$ implies that the fourth factor $x + y - z$ is positive. Also, a side of a triangle cannot be zero, from which we get $x + y + z > 0$, $-x + y + z > 0$, $x - y + z > 0$, and hence $f(x, y, z) > 0$.

Conversely, if $f(x, y, z) > 0$, then the four factors must be positive, so $x, y$, and $z$ are positive and the triangle inequality $x + y > z$ holds. To construct the triangle, we may draw two circles of radii $x$ and $y$ whose centers $Y$, $X$ are a distance $z$ apart. Because each circle passes both inside and outside the other, the circles intersect at two points. Let $Z$ be one. Then $X Y Z$ is the desired triangle.