One, (20 points) Given real numbers $a, b, c, d$ satisfy $a+b+c+d=4$, $ab+ac+ad+bc+bd+cd=-\frac{14}{3}$. Find the maximum and minimum values of $b+c+d$.

保留源文本的换行和格式，直接输出翻译结果。
Let's assume $b+c+d=t$. Then
$$
\begin{array}{l}
a=4-t, \\
b c+b d+c d=-\frac{14}{3}-a t \\
=-\frac{14}{3}-(4-t) t=t^{2}-4 t-\frac{14}{3} .
\end{array}
$$

Notice that
$$
\begin{array}{l}
2(b+c+d)^{2}-6(b c+b d+c d) \\
=(b-c)^{2}+(b-d)^{2}+(c-d)^{2} \geqslant 0,
\end{array}
$$

i.e., $2 t^{2}-6\left(t^{2}-4 t-\frac{14}{3}\right) \geqslant 0$.
Thus, $t^{2}-6 t-7 \leqslant 0$.
Solving this, we get $-1 \leqslant t \leqslant 7$.
When $t=7$, $a=-3$, at this point, equation (1) holds with equality, yielding $b=c=d=\frac{7}{3}$, which satisfies the given conditions;

When $t=-1$, $a=5$, at this point, equation (1) holds with equality, yielding $b=c=d=-\frac{1}{3}$, which satisfies the given conditions.
Therefore, the maximum value of $b+c+d$ is 7, and the minimum value is -1.