28. Let $a=\frac{x}{z}, b=\frac{y}{x}, c=\frac{z}{y}$, then $\frac{a}{b^{2}(c+1)}+\frac{b}{c^{2}(a+1)}+\frac{c}{a^{2}(b+1)} \geqslant \frac{3}{2} \Leftrightarrow$ $\frac{x^{3}}{y z(y+z)}+\frac{y^{3}}{z x(z+x)}+\frac{z^{3}}{x y(x+y)} \geqslant \frac{3}{2} \Leftrightarrow \frac{x^{4}}{x y z(y+z)}+\frac{y^{4}}{x y z(z+x)}+$ $\frac{z^{4}}{x y z(x+y)} \geqslant \frac{3}{2}$.
By the Cauchy-Schwarz inequality, we have
$$\left(\frac{x^{4}}{y+z}+\frac{y^{4}}{z+x}+\frac{z^{4}}{x+y}\right)[(y+z)+(z+x)+(x+y)] \geqslant\left(x^{2}+y^{2}+z^{2}\right)^{2}$$

Thus, it suffices to prove
$$\left(x^{2}+y^{2}+z^{2}\right)^{2} \geqslant 3 x y z(x+y+z)$$

In fact,
$$\begin{aligned} \left(x^{2}+y^{2}+z^{2}\right)^{2} \geqslant & \frac{1}{3}(x+y+z)^{2}\left(x^{2}+y^{2}+z^{2}\right)= \\ & \frac{1}{3}(x+y+z)(x+y+z)\left(x^{2}+y^{2}+z^{2}\right) \geqslant \\ & \frac{1}{3}(x+y+z) \cdot 3 \sqrt[3]{x y z} \cdot 3 \sqrt[3]{(x y z)^{2}}= \\ & 3 x y z(x+y+z) \end{aligned}$$