$9 \cdot 40$ Find the smallest real number $a$, such that for any non-negative real numbers $x$, $y$, $z$ whose sum is 1, we have
$$a\left(x^{2}+y^{2}+z^{2}\right)+x y z \geqslant \frac{a}{3}+\frac{1}{27} .$$
Let $I=a\left(x^{2}+y^{2}+z^{2}\right)+x y z$. Without loss of generality, assume $x \leqslant y \leqslant z$, and let $x=\frac{1}{3}+\delta_{1}, y=\frac{1}{3}+\delta_{2}, z=\frac{1}{3}+\delta_{3}$, then
$$-\frac{1}{3} \leqslant \delta_{1} \leqslant 0,0 \leqslant \delta_{3} \leqslant \frac{2}{3}, \delta_{1}+\delta_{2}+\delta_{3}=0$$

Express $I$ using the new variables $\delta_{1}, \delta_{2}, \delta_{3}$, then
$$I=\frac{a}{3}+\frac{1}{27}+a\left(\delta_{1}^{2}+\delta_{2}^{2}+\delta_{3}^{2}\right)+\frac{1}{3}\left(\delta_{1} \delta_{2}+\delta_{2} \delta_{3}+\delta_{3} \delta_{1}\right)+$$
$\delta_{1} \delta_{2} \delta_{3}$
From (1), we know $\delta_{1} \delta_{2}+\delta_{2} \delta_{3}+\delta_{3} \delta_{1}=-\frac{1}{2}\left(\delta_{1}^{2}+\delta_{2}^{2}+\delta_{3}^{2}\right)$, thus
$$I=\frac{a}{3}+\frac{1}{27}+\left(a-\frac{1}{6}\right)\left(\delta_{1}^{2}+\delta_{2}^{2}+\delta_{3}^{2}\right)+\delta_{1} \delta_{2} \delta_{3}$$

If $a<\frac{2}{9}$, take $x=0, y=z=\frac{1}{2}$, i.e., $\delta_{1}=-\frac{1}{3}, \delta_{2}=\delta_{3}=\frac{1}{6}$, since
$$\begin{array}{l}
\left(a-\frac{1}{6}\right)\left(\delta_{1}^{2}+\delta_{2}^{2}+\delta_{3}^{2}\right)+\delta_{1} \delta_{2} \delta_{3} \\
<\frac{1}{18} \cdot\left(\frac{1}{9}+\frac{2}{36}\right)-\frac{1}{3 \cdot 36}=0,
\end{array}$$

Therefore, at this time, $I<\frac{a}{3}+\frac{1}{27}$.
Thus, the required $a \geqslant \frac{2}{9}$. When $a \geqslant \frac{2}{9}$, if $\delta_{2} \leqslant 0$, then from (1) we know $\delta_{1} \delta_{2} \delta_{3} \geqslant 0$, which immediately gives $I \geqslant \frac{a}{3}+\frac{1}{27}$.
Assume $\delta_{2} \geqslant 0$, by $\delta_{1}=-\left(\delta_{2}+\delta_{3}\right)$, then
$$\begin{aligned}
I= & \frac{a}{3}+\frac{1}{27}+\left(a-\frac{1}{6}\right)\left(2 \delta_{2}^{2}+2 \delta_{3}^{2}+2 \delta_{2} \delta_{3}\right)-\left(\delta_{2}+\right. \\
& \left.\delta_{3}\right) \delta_{2} \delta_{3} \\
= & \frac{a}{3}+\frac{1}{27}+2\left(a-\frac{1}{6}\right)\left(\delta_{2}-\delta_{3}\right)^{2}+ \\
& {\left[6\left(a-\frac{1}{6}\right)-\left(\delta_{2}+\delta_{3}\right)\right] \delta_{2} \delta_{3} . }
\end{aligned}$$

From $a \geqslant \frac{2}{9}$, we get $a-\frac{1}{6} \geqslant \frac{1}{18}$, and from (1) we get $0 \leqslant \delta_{2}+\delta_{3}=-\delta_{1} \leqslant \frac{1}{3}$, so $I \geqslant \frac{a}{3}+\frac{1}{27}$.

In summary, the smallest real number $a=\frac{2}{9}$.