6. Given $x, y \in \mathbf{R}$. If
$$
a(x+y) \leqslant 1+x^{2}+x y+y^{2}
$$

always holds, then the range of values for $a$ is $\qquad$ _.
6. $[-\sqrt{3}, \sqrt{3}]$.

Notice that
$$
\begin{array}{l}
a(x+y) \leqslant 1+x^{2}+x y+y^{2} \\
=1+\frac{3}{4}(x+y)^{2}+\frac{1}{4}(x-y)^{2} .
\end{array}
$$

If $x+y=0$, it is clear that $a \in \mathbf{R}$.
If $x+y>0$, then
$$
a \leqslant \frac{1}{x+y}+\frac{3}{4}(x+y)+\frac{1}{4} \cdot \frac{(x-y)^{2}}{x+y}
$$

always holds.
$$
\begin{array}{l}
\text { Also, } \frac{1}{x+y}+\frac{3}{4}(x+y)+\frac{1}{4} \cdot \frac{(x-y)^{2}}{x+y} \\
\geqslant \frac{1}{x+y}+\frac{3}{4}(x+y) \geqslant \sqrt{3},
\end{array}
$$

with equality if and only if $x=y=\frac{1}{\sqrt{3}}$.
Thus, $a \leqslant \sqrt{3}$.
If $x+y<0$, similarly, we get $a \geqslant-\sqrt{3}$.
In summary, $-\sqrt{3} \leqslant a \leqslant \sqrt{3}$.