7. Find the maximum value of the positive real number $M$ such that for all $x, y, z \in \mathbf{R}$, the inequality $x^{4}+y^{4}+z^{4}+$ $x y z(x+y+z) \geqslant M(x y+y z+z x)^{2}$ holds. (2004 Greek Mathematical Olympiad)
7. When $x=y=z$, the inequality becomes $6 x^{4} \geqslant M \cdot 9 x^{4}, x \in \mathbf{R}$, i.e., $M \leqslant \frac{2}{3}$. We will now prove that the maximum value of $M$ is $\frac{2}{3}$.
$$\begin{array}{l}
\quad \text { For } x, y, z \in \mathbf{R}, x^{4}+y^{4}+z^{4}+x y z(x+y+z) \geqslant \frac{2}{3}(x y+y z+z x)^{2} \Leftrightarrow \\
3\left(x^{4}+y^{4}+z^{4}\right)+3 x y z(x+y+z) \geqslant 2(x y+y z+z x)^{2} .
\end{array}$$

Since $x^{4}+y^{4}+z^{4} \geqslant x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2}$, it suffices to prove that $3\left(x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2}\right)+$ $3 x y z(x+y+z) \geqslant 2(x y+y z+z x)^{2}$.
That is,
$$x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2} \geqslant x y z(x+y+z)$$

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