88 = If $a, b$ are positive numbers, then $\sqrt[3]{-\frac{a^{3}+b^{3}}{2}} \leqslant \frac{a^{2}+b^{2}}{a+b}$. (Proof see Question 60), thus
$$\begin{array}{l} \frac{\left(-1+a^{2}\right)\left(1+b^{2}\right):\left(1+c^{2}\right)}{(1+a)(1+b)=(1+c)} \geqslant \sqrt{\frac{1+a^{3}}{2}} \cdot \sqrt[3]{\frac{1+b^{3}}{2}} \cdot \sqrt[3]{\frac{1+c^{3}}{2}}= \\ \frac{1}{2} \cdot \sqrt[3]{\left(1+a^{3}\right)\left(1+b^{3}\right)\left(1+c^{3}\right) *} \end{array}$$

By the AM-GM inequality, we get
$$\begin{aligned} \left(1+a^{3}\right)\left(1+b^{3}\right)-\left(1+c^{3}\right)= & 1+\left(a^{3}+b^{3}+c^{3}\right)+\left(a^{3} b^{3}+b^{3} c^{3}+c^{3} a^{3}\right)+a^{3} b^{3} c^{3} \geqslant \\ & 1+3 a b c+3 a^{2} b^{2} c^{2}+a^{3} b^{3} c^{3}=(1+a b c)^{3-} \end{aligned}$$

Therefore,
$$\frac{\left(1+a^{2}\right)\left(-1+b^{2}\right)\left(F+c^{2}\right)}{(1+a)(1+b)(1+c)} \geqslant-\frac{1}{2}(1+a b c)$$