Solve: By Cauchy-Schwarz inequality, we have $\left(1^{2}+1^{2}+1^{2}\right)\left(x^{2}+y^{2}+\right.$ $\left.z^{2}\right) \geqslant(x+y+z)^{2}$, and $x+y+z=3, x^{2}+y^{2}+z^{2}=$ 3, so the equality of the above inequality holds, thus $x=y=z=1$, which also satisfies $x^{5}+y^{5}+z^{5}=3$, hence the solution to the original system of equations is $x=y=$

Reference [1] gives the following inequality:
Given $x, y, z \in \mathbf{R}^{+}$, and $x+y+z=1$, then
$$\left(\frac{1}{x}-x\right)\left(\frac{1}{y}-y\right)\left(\frac{1}{z}-z\right) \geqslant\left(\frac{8}{3}\right)^{3}$$

The original text proves inequality (1) using advanced methods, but the process is rather complex. Below, the author provides a simple elementary proof of the inequality.

Proof $\frac{1}{x}-x=\frac{1-x^{2}}{x}=\frac{(1+x)(1-x)}{x}=(y$ $+z) \cdot \frac{1+x}{x}$, similarly, $\frac{1}{y}-y=(x+z) \cdot \frac{1+y}{y}, \frac{1}{z}-z$ $=(x+y) \cdot \frac{1+z}{z}$

Multiplying the three equations, we get: $\left(\frac{1}{x}-x\right)\left(\frac{1}{y}-y\right)\left(\frac{1}{z}-z\right)=(x+$ $y)(y+z)(z+x) \cdot \frac{(1+x)(1+y)(1+z)}{x y z}$, using the identity $(x+y)(y+z)(z+x)=(x+y+z)(x y+y z$ $+z x)-x y z$, hence $\left(\frac{1}{x}-x\right)\left(\frac{1}{y}-y\right)\left(\frac{1}{z}-z\right)=[(x y$ $+y z+z x)-x y z] \cdot \frac{(x y+y z+z x)+x y z+2}{x y z}$ $=$ $\frac{(x y+y z+z x)^{2}-(x y z)^{2}+2(x y+y z+z x)-2 x y z}{x y z}$ $=\left[x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2}+2 x y z(x+y+z)-(x y z)^{2}\right.$ $+2(x y+y z+z x)-2 x y z] / x y z$ $\geqslant \frac{x y z(x+y+z)+2(x y+y z+z x)-(x y z)^{2}}{x y z}$ $=1+2\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)-x y z$ $\geqslant 1+2 \times 9-\frac{1}{27}=19-\frac{1}{27}=\frac{512}{27}=\left(\frac{8}{3}\right)^{3}$.

Equality holds if and only if $x=y=z=\frac{1}{3}$.