Four, (20 points) If real numbers $a, b, c$ satisfy $a^{2}+b^{2} \leqslant c \leqslant 1$, find the maximum and minimum values of $a+b+c$.


Let $a=r \cos \theta, b=r \sin \theta(\theta \in \mathbf{R})$. Then $r=\sqrt{a^{2}+b^{2}}(0 \leqslant r \leqslant \sqrt{c} \leqslant 1)$.
Thus, $a+b+c=r(\cos \theta+\sin \theta)+c$
$$
=\sqrt{2} r \sin \left(\theta+\frac{\pi}{4}\right)+c \text {. }
$$

By $-1 \leqslant \sin \left(\theta+\frac{\pi}{4}\right) \leqslant 1$, we know
$$
\begin{array}{l}
a+b+c \leqslant \sqrt{2} r+c \leqslant 1+\sqrt{2} . \\
\text { Also, } a+b+c \geqslant-\sqrt{2} r+c \geqslant-\sqrt{2 c}+c \\
=\left(c-\frac{\sqrt{2}}{2}\right)^{2}-\frac{1}{2} \geqslant-\frac{1}{2},
\end{array}
$$

Thus, $-\frac{1}{2} \leqslant a+b+c \leqslant 1+\sqrt{2}$.
When $a=b=\frac{\sqrt{2}}{2}, c=1$, we have $a^{2}+b^{2}=c=1$, at this time,
$$
(a+b+c)_{\max }=1+\sqrt{2} \text {; }
$$

When $a=b=-\frac{1}{2}, c=\frac{1}{2}$, we have $a^{2}+b^{2}=c=\frac{1}{2}$, at this time,
$$
(a+b+c)_{\text {min }}=-\frac{1}{2} \text {. }
$$