Example 1.33
$$\begin{array}{c}
\text { Let } x, y, z \in \mathbf{R}^{+} \text {, and }(x+y+z)^{3}=32 x y z \text {, find } \\
p=\frac{x^{4}+y^{4}+z^{4}}{(x+y+z)^{4}}
\end{array}$$

the maximum and minimum values.
Let $\sum x=1$, then $x y z=\frac{1}{32}$, denote $s=x y+y z+z x$, then

that is
$$\begin{array}{c}
p=2\left(\sum x y-1\right)^{2}-\frac{7}{8} \\
p=2(s-1)^{2}-\frac{7}{8}
\end{array}$$

On the other hand
$$(1-x)^{2}=(y+z)^{2} \geqslant 4 y z=\frac{1}{8 x} \Rightarrow \frac{3-\sqrt{5}}{4} \leqslant x \leqslant \frac{1}{2}$$

Similarly, we have
$$\begin{array}{l}
\frac{3-\sqrt{5}}{4} \leqslant x, y, z \leqslant \frac{1}{2} \Rightarrow \\
\left(x-\frac{3-\sqrt{5}}{4}\right)\left(y-\frac{3-\sqrt{5}}{4}\right)\left(z-\frac{3-\sqrt{5}}{4}\right) \geqslant 0
\end{array}$$

and $\left(x-\frac{1}{2}\right)\left(y-\frac{1}{2}\right)\left(z-\frac{1}{2}\right) \leqslant 0$ which gives
$$\frac{5}{16} \leqslant s \leqslant \frac{5 \sqrt{5}-1}{32}$$

Substituting into $p$ we get
$$\frac{383-165 \sqrt{5}}{256} \leqslant p \leqslant \frac{9}{128}$$

When $x=y=\frac{1}{4}, z=\frac{1}{2}$, $p$ reaches its maximum value $\frac{9}{128}$.
When $x=y=\frac{\sqrt{5}+1}{8}, z=\frac{3-\sqrt{5}}{4}$, $p$ reaches its minimum value $\frac{383-165 \sqrt{5}}{256}$.