55. Let $x, y, z$ be positive numbers, and $(x+y+z)^{3}=32 x y z$, find the maximum and minimum values of $\frac{x^{4}+y^{4}+z^{4}}{(x+y+z)^{4}}$. (2004 Vietnam Mathematical Olympiad)
55. Let's assume $x+y+z=4, xyz=2$, so we only need to find the maximum and minimum values of $\frac{x^{4}+y^{4}+z^{4}}{4^{4}}$. Now we have $x^{4}+y^{4}+z^{4}=\left(x^{2}+y^{2}+z^{2}\right)^{2}-2\left(x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2}\right)=$ $[16-2(xy+yz+zx)]^{2}-2(xy+yz+zx)^{2}+4xyz(x+y+z)=a^{2}-64a+288$, where $a=xy+yz+zx$. Since $y+z=4-x, yz=\frac{2}{x}$, we have $(4-x)^{-2} \geqslant \frac{8}{x}$, which implies $3-\sqrt{5} \leqslant x \leqslant 2$. By symmetry, $x, y, z \in[3-\sqrt{5}, 2]$, so $(x-2)(y-2)(z-2) \leqslant 0, (x-3+\sqrt{5})(y-3+\sqrt{5})(z-3+\sqrt{5}) \geqslant 0$, thus $a \in\left[5, \frac{5 \sqrt{5} + 1}{2}\right]$. Since $\frac{x^{4}+y^{4}+z^{4}}{4^{4}}=\frac{(a-16)^{2}-112}{4^{4}}$, the minimum value of $\frac{x^{4}+y^{4}+z^{4}}{4^{4}}$ is $\frac{383-165 \sqrt{5}}{256}$, and the maximum value is $\frac{9}{128}$. The minimum value $\frac{383-165 \sqrt{5}}{256}$ is achieved when $(x, y, z)=\left(3-\sqrt{5}, \frac{\sqrt{5}+1}{2}, \frac{\sqrt{5}+1}{2}\right)$ or permutations thereof, and the maximum value $\frac{9}{128}$ is achieved when $(x, y, z)=(2,1,1)$ or permutations thereof.