8. Given $a, b, c, d \in \mathbf{R}$, and $a^{2}+b^{2}+c^{2}+d^{2}$ $=1$. Then the maximum value of $a b+b c+c d+d a+a c+4 b d$ is $\qquad$
8. $\frac{5}{2}$.

From the condition, we have
$$
\begin{array}{l}
a^{2}+b^{2}+c^{2}+d^{2} \\
=\frac{1}{10}\left(10 a^{2}+10 b^{2}+10 c^{2}+10 d^{2}\right) \\
=\frac{1}{10}\left(\left(4 a^{2}+b^{2}\right)+\left(2 a^{2}+2 c^{2}\right)+\left(4 a^{2}+d^{2}\right)+\right. \\
\left.\left(b^{2}+4 c^{2}\right)+\left(4 c^{2}+d^{2}\right)+\left(8 b^{2}+8 d^{2}\right)\right) \\
\geqslant \frac{1}{10}(4 a b+4 a c+4 a d+4 b c+4 c d+16 b d) \\
\Rightarrow a b+b c+c d+d a+a c+4 b d \leqslant \frac{5}{2} \text {, } \\
\end{array}
$$

Equality holds if and only if $a=c=\frac{\sqrt{10}}{10}, b=d=\frac{2 \sqrt{10}}{10}$.