Three, (50 points) Given non-negative real numbers $a, b, c, d$ satisfying $a+b+c+d=4$. Find the minimum value of $\sum \frac{b+3}{a^{2}+4}$, where “$\sum$” denotes the cyclic sum.
$$
\frac{1}{a^{2}+4} \geqslant \frac{4-a}{16} \Leftrightarrow a(a-2)^{2} \geqslant 0 \text {. }
$$

Obviously true.
$$
\begin{array}{l}
\text { Then } \sum \frac{b+3}{a^{2}+4} \geqslant \sum \frac{(b+3)(4-a)}{16} \\
=\frac{1}{16}\left(\sum a-\sum a b\right)+3 . \\
\text { Also } \sum a b=(a+c)(b+d) \\
\leqslant \frac{1}{4}\left(\sum a\right)^{2}=4, \\
\text { hence } \sum \frac{b+3}{a^{2}+4} \geqslant 3 .
\end{array}
$$

When $a=b=2, c=d=0$, the equality holds.
Therefore, the minimum value of $\sum \frac{b+3}{a^{2}+4}$ is 3.