Proof Let
$$a=\frac{x}{y}, b=\frac{y}{z}, c=\frac{z}{x}\left(x, y, z \in \mathbf{R}_{+}\right) \text {. }$$

Then the original inequality
$$\begin{array}{l} \Leftrightarrow \sum \frac{(x-y)(z+x)}{y(x+y+z)} \geqslant 0 \\ \Leftrightarrow \sum \frac{z x+x^{2}-y z-x y}{y} \geqslant 0 \\ \Leftrightarrow \sum\left(\frac{z x}{y}+\frac{x^{2}}{y}\right) \geqslant 2 \sum x, \end{array}$$

where, “ $\sum$ ” denotes the cyclic sum.
By the AM-GM inequality, we have
$$\begin{array}{l} \sum \frac{z x}{y}=\frac{1}{2} \sum\left(\frac{z x}{y}+\frac{x y}{z}\right) \geqslant \sum x, \\ \sum \frac{z^{2}}{x}=\sum\left(\frac{z^{2}}{x}+x\right)-\sum x \geqslant x+y+z . \end{array}$$

Adding equations (2) and (3) yields equation (1).
Thus, the original inequality holds.