15 Find the maximum value of the positive real number $A$ such that for any real numbers $x, y, z$, the inequality
$$
x^{4}+y^{4}+z^{4}+x^{2} y z+x y^{2} z+x y z^{2}-A(x y+y z+z x)^{2} \geqslant 0
$$

holds.
15 When $x=y=z$, the inequality transforms into $6 x^{4} \geqslant A \cdot 9 x^{4}, x \in \mathbf{R}$, which means $A \leqslant \frac{2}{3}$.
Next, we prove: The maximum value of $A$ is $\frac{2}{3}$.
For $x, y, z \in \mathbf{R}$,
$$
x^{4}+y^{4}+z^{4}+x y z(x+y+z) \geqslant \frac{2}{3}(x y+y z+z x)^{2} .
$$

It suffices to prove
$$
3\left(x^{4}+y^{4}+z^{4}\right)+3 x y z(x+y+z) \geqslant 2(x y+y z+z x)^{2} .
$$

Since
$$
x^{4}+y^{4}+z^{4} \geqslant x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2},
$$

it is sufficient to prove
$$
3\left(x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2}\right)+3 x y z(x+y+z) \geqslant 2(x y+y z+z x)^{2} .
$$

That is, to prove
$$
x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2} \geqslant x y z(x+y+z) .
$$

Since
$$
(x y-y z)^{2}+(y z-z x)^{2}+(z x-x y)^{2} \geqslant 0,
$$

the conclusion holds.