For $x \in \mathbf{R}^{+}, x^{2}-1$ and $x^{3}-1$ have the same sign, so
$$\left(x^{2}-1\right)\left(x^{3}-1\right) \geqslant 0$$

which implies $x^{5}-x^{2}+3 \geqslant x^{3}+2$.
Thus, $\left(a^{5}-a^{2}+3\right)\left(b^{5}-b^{2}+3\right)\left(c^{5}-c^{2}+3\right)$
$$\geqslant\left(a^{3}+2\right)\left(b^{3}+2\right)\left(c^{3}+2\right)$$

By (7) (Hölder's inequality), we have
$$\begin{aligned} & \left(a^{3}+2\right)\left(b^{3}+2\right)\left(c^{3}+2\right) \\ = & \left(a^{3}+1+1\right)\left(1+b^{3}+1\right)\left(1+1+c^{3}\right) \\ \geqslant & (a+b+c)^{3} \end{aligned}$$

Thus, the proposition is proved.