89. Simplified Proof
$$\begin{array}{l} \text { Original } \Leftrightarrow \sum(1+b)(1+c)\left(1+a^{2}\right) \leqslant 3 \prod\left(1+a^{2}\right) \Leftrightarrow \\ 12-8 \sum b c+3\left(\sum b c\right)^{2}-18 a b c+3(a b c)^{2} \geqslant 0 \end{array}$$

Let $s_{1}=\sum a, s_{2}=\sum b c, s_{3}=a b c$, noting that $\sum a=3$, homogenizing equation (1), we get
$$\begin{array}{l} 12 \cdot\left(\frac{\sum a}{3}\right)^{6}-8\left(\frac{\sum a}{3}\right)^{4} \cdot \sum b c+3\left(\frac{\sum a}{3}\right)^{2} \cdot\left(\sum b c\right)^{2}- \\ 18\left(\frac{\sum a}{3}\right)^{3} \cdot a b c+3(a b c)^{3} \geqslant 0 \Leftrightarrow \\ 4 s_{1}^{4}-24 s_{1}^{4} s_{2}+81 s_{1}^{2} s_{2}^{2}-162 s_{1}^{3} s_{3}+729 s_{3}^{2} \geqslant 0 \end{array}$$

Therefore, to prove the original inequality, it suffices to prove equation (2). For this, let $s_{1}=1$ in equation (2), we get
$$4-24 s_{2}+81 s_{2}^{2}-162 s_{3}+729 s_{3}^{2} \geqslant 0$$

Thus, it suffices to prove equation (3), where $s_{1}=\sum a=1$.
$$\begin{array}{c} \text { Let } \omega=\sqrt{1-3 \sum b c} \text {, i.e., } \sum b c=\frac{1-\omega^{2}}{3}, 0 \leqslant x \leqslant 1 \text {, then } \\ s_{3} \leqslant \frac{1-3 \omega^{2}+2 \omega^{3}}{27} \end{array}$$
(Refer to the corollary 3 of theorem 2 in the appendix of the chapter "Application of Basic Inequalities to Prove Inequalities") Therefore, we have

The left side of equation (3) $\geqslant 4-24 \cdot \frac{1-\omega^{2}}{3}+81 \cdot\left(\frac{1-\omega^{2}}{3}\right)^{2}-162 \cdot \frac{1-3 \omega^{2}+2 \omega^{3}}{27}+$
$$\begin{array}{l} 729 \cdot\left(\frac{1-3 \omega^{2}+2 \omega^{3}}{27}\right)^{2}= \\ 4 \omega^{2}\left(2-2 \omega+3 \omega^{2}-3 \omega^{3}+\omega^{4}\right)= \\ 4 \omega^{2}\left[(1-\omega)+1-\omega(1-\omega)^{3}\right] \geqslant 0 \end{array}$$
(Noting that $0 \leqslant \omega \leqslant 1$) Equation (3) holds, thus equation (1) holds, and the original inequality is proved.
Note: (1) From the original inequality, we can derive: If $a, b, c \in \overline{\mathbf{R}^{-}}$, and $a+b+c=3$, then
$$\Pi\left(1+a^{2}\right) \geqslant \Pi(1+a)$$

Equality holds if and only if $a=b=c=1$.
(2) Conjecture: Suppose $a_{i} \in \overline{\mathbf{R}^{-}}, i=1,2, \cdots, n, m, n$ are positive integers, and $a_{1}+a_{2}+\cdots+$ $a_{n}=n$, then
$$\prod_{i=1}^{n}\left(1+a_{i}^{m}\right) \geqslant \prod_{i=1}^{n}\left(1+a_{i}^{m-1}\right)$$

Equality holds if and only if $a_{1}=a_{2}=\cdots=a_{n}=1$.
(3) Further conjecture: Suppose $a_{i} \in \overline{\mathbf{R}^{-}}, i=1,2, \cdots, n, m, n$ are positive integers, and $\sum_{i=1}^{n} a_{i}=n$, then
$$\sum_{i=1}^{n} \frac{1+a_{i}^{m}}{1+a_{i}^{m-1}} \leqslant n \prod_{i=1}^{n} \frac{1+a_{i}^{m}}{1+a_{i}^{m-1}}$$

Equality holds if and only if $a_{1}=a_{2}=\cdots=a_{n}=1$.
(4) Conjecture: Suppose $a_{i} \in \overline{\mathbf{R}^{-}}$, and $\sum_{i=1}^{n} a_{i}=n, m, n$ are positive integers, and $m \geqslant n$, then
$$\prod_{i=1}^{n}\left(1+a_{i}^{m}\right) \geqslant 2^{n}$$

Equality holds if and only if $a_{1}=a_{2}=\cdots=a_{n}=1$.