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

holds.
$$
a_{2}=1, a_{3}=\frac{5}{2}, a_{4}=\frac{9}{2}, \cdots,
$$

Guess: $a_{n}=\frac{(n-1)(n+2)}{4}$, then prove it using mathematical induction.
15. When $x=y=z$, the inequality transforms to
$$
6 x^{4} \geqslant A \cdot 9 x^{4} (x \in \mathbf{R}),
$$

which means $A \leqslant \frac{2}{3}$.
Next, we prove: the maximum value of $A$ is $\frac{2}{3}$.
For $x, y, z \in \mathbf{R}$,
$$
\begin{array}{l}
x^{4}+y^{4}+z^{4}+x y z(x+y+z) \\
\geqslant \frac{2}{3}(x y+y z+z x)^{2},
\end{array}
$$

which is
$$
\begin{array}{l}
3\left(x^{4}+y^{4}+z^{4}\right)+3 x y z(x+y+z) \\
\geqslant 2(x y+y z+z x)^{2}. \\
\text { Also, } x^{4}+y^{4}+z^{4} \geqslant x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2}, \text { so we only need to } \\
3\left(x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2}\right)+3 x y z(x+y+z) \\
\geqslant 2(x y+y z+z x)^{2},
\end{array}
$$

Also, $x^{4}+y^{4}+z^{4} \geqslant x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2}$, so we only need to prove:
$$
\begin{array}{l}
3\left(x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2}\right)+3 x y z(x+y+z) \\
\geqslant 2(x y+y z+z x)^{2},
\end{array}
$$

which is $x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2} \geqslant x y z(x+y+z)$.
Since $(x y-y z)^{2}+(y z-z x)^{2}+(z x-x y)^{2} \geqslant 0$, the conclusion holds.