6. Given $x^{2}+y^{2}+z^{2}=3$. Then the minimum value of $x y+y z+z x$ is $\qquad$ .
6. $-\frac{3}{2}$.
$$
\begin{array}{l}
\text { From }(x+y+z)^{2} \\
=x^{2}+y^{2}+z^{2}+2(x y+y z+z x) \geqslant 0,
\end{array}
$$

we know that the minimum value is $-\frac{3}{2}$, which is achieved when $x+y+z=0$ and $x^{2}+y^{2}+z^{2}=3$. For example, $(x, y, z)=\left(\sqrt{\frac{3}{2}},-\sqrt{\frac{3}{2}}, 0\right)$.