Clothing 193 Let $a, b, c \in \mathbf{R}_{+}$, and satisfy
$$
\begin{array}{l}
a^{2}+b^{2}+c^{2}+a b c=4 . \\
\text { Let } m=a^{3}+b^{3}+c^{3}, n=1+2 a b c, s=a+b+c .
\end{array}
$$

Try to compare the sizes of $m, n, s$, and find the maximum and minimum values of $m, n, s$ respectively.
Solution: (1) Obviously, when $a=b=c=1$, $a^{2}+b^{2}+c^{2}+a b c=4$, and $m=n=s=3$.
(2) It can be found that $(a, b, c)=\left(\frac{7}{4}, \frac{1}{2}, \frac{1}{2}\right)$ satisfies $a^{2}+b^{2}+c^{2}+a b c=4$.
For this, it is only necessary to take $b=c=\frac{1}{2}$, at this time,
$$
4=a^{2}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4} a,
$$

which is $4 a^{2}+a-14=0$.
Solving gives $a_{1}=\frac{7}{4}, a_{2}=-2$ (discard).
(3) Guess: $m=a^{3}+b^{3}+c^{3} \geqslant 3$
$$
\geqslant s=a+b+c \geqslant 1+2 a b c=n .
$$

For $(a, b, c)=\left(\frac{7}{4}, \frac{1}{2}, \frac{1}{2}\right)$, it is easy to know $m=\left(\frac{7}{4}\right)^{3}+\left(\frac{1}{2}\right)^{3}+\left(\frac{1}{2}\right)^{3}=5 \frac{39}{64}$, $s=\frac{7}{4}+\frac{1}{2}+\frac{1}{2}=2 \frac{3}{4}$,
$$
n=1+2 \times \frac{7}{4} \times \frac{1}{2} \times \frac{1}{2}=1 \frac{7}{8} \text {. }
$$

Obviously, $m>3>s>n$.
Below is a brief proof:
$$
a^{3}+b^{3}+c^{3} \geqslant 3 \geqslant a+b+c \geqslant 1+2 a b c .
$$

First, prove: $a^{3}+b^{3}+c^{3} \geqslant 3$.
According to the problem, we know
$$
\begin{array}{l} 
4=a^{2}+b^{2}+c^{2}+a b c \geqslant 4 \sqrt[4]{a^{2} \cdot b^{2} \cdot c^{2} \cdot a b c} \\
=4 \sqrt[4]{(a b c)^{3}}
\end{array}
$$

Thus, $a b c \leqslant 1$. And at this time,
$$
4=a^{2}+b^{2}+c^{2}+a b c \leqslant a^{2}+b^{2}+c^{2}+1,
$$

So, $a^{2}+b^{2}+c^{2} \geqslant 3$.
It is easy to know $\left(a^{2}+b^{2}+c^{2}\right)^{2} \geqslant 3\left(a^{2}+b^{2}+c^{2}\right)$
$\geqslant(a+b+c)^{2}$,
Then, $a^{2}+b^{2}+c^{2} \geqslant a+b+c$.
Therefore, $a^{3}+b^{3}+c^{3}=\frac{a^{4}}{a}+\frac{b^{4}}{b}+\frac{c^{4}}{c}$
$$
\geqslant \frac{\left(a^{2}+b^{2}+c^{2}\right)^{2}}{a+b+c} \geqslant \frac{\left(a^{2}+b^{2}+c^{2}\right)(a+b+c)}{a+b+c}
$$
$\geqslant 3$.
Next, prove: $3 \geqslant a+b+c$.
Considering
$$
\begin{aligned}
4 & =a^{2}+b^{2}+c^{2}+a b c \geqslant a^{2}+2 b c+a b c \\
& =a^{2}+b c(2+a),
\end{aligned}
$$

which is $4-a^{2} \geqslant b c(2+a)$.
So, $2-a \geqslant b c$, that is
$2 \geqslant a+b c$.
Note that $a, b, c$ cannot all be greater than 1, nor can they all be less than 1. They can only be two numbers not less than 1 (at this time, the third number is not greater than 1), or two numbers not greater than 1 (at this time, the third number is not less than 1).

Without loss of generality, assume these two numbers are $b, c$. At this time, $(b-1)(c-1) \geqslant$ 0, then $b c-b-c+1 \geqslant 0$, that is
$b c+1 \geqslant b+c$.
(1) + (2) gives $(b c+1)+2 \geqslant(b+c)+(a+b c)$.
So, $3 \geqslant a+b+c$.
Finally, prove: $a+b+c \geqslant 1+2 a b c$.
Considering that it has been proven $2 \geqslant a+b c$, hence $2 a \geqslant a^{2}+a b c$.
Similarly, $2 b \geqslant b^{2}+a b c, 2 c \geqslant c^{2}+a b c$.
Adding the three equations gives
$$
\begin{array}{l}
2(a+b+c) \geqslant\left(a^{2}+b^{2}+c^{2}\right)+3 a b c \\
\geqslant 3+3 a b c \geqslant 2+4 a b c=2(1+2 a b c) .
\end{array}
$$

Thus, $a+b+c \geqslant 1+2 a b c$.
In summary, $m \geqslant 3 \geqslant s \geqslant n$.
When and only when $a=b=c=1$, $m=a^{3}+b^{3}+c^{3}$ has the minimum value $3, s=a+b+c$ and $n=1+2 a b c$ have the maximum value 3.
(Jianbin Sun, Fujian Yongchun County Science and Technology Commission, 362600)