First, prove
$$\frac{1}{1+a+a^{2}+a^{3}}+\frac{1}{1+b+b^{2}+b^{3}} \geqslant \frac{1}{1+\sqrt{(a b)^{3}}}$$

Equation (24) is equivalent to, for $x, y \in \mathbf{R}^{+}$, we have
$$\frac{1}{1+x^{2}+x^{4}+x^{6}}+\frac{1}{1+y^{2}+y^{4}+y^{6}} \geqslant \frac{1}{1+x^{3} y^{3}}$$

Equation $(※) \Leftrightarrow\left(1+y^{2}+y^{4}+y^{6}+1+x^{2}+x^{4}+x^{6}\right)\left(1+x^{3} y^{3}\right) \geqslant$ $\left(1+y^{2}+y^{4}+y^{6}\right)\left(1+x^{2}+x^{4}+x^{6}\right) \Leftrightarrow$ $1+2 x^{3} y^{3}+x^{3} y^{3}\left(x^{2}+y^{2}\right)+x^{3} y^{3}\left(x^{4}+y^{4}\right)+x^{3} y^{3}\left(x^{6}+y^{6}\right) \geqslant$ $x^{2} y^{2}+x^{4} y^{4}+x^{6} y^{6}+x^{2} y^{2}\left(x^{2}+y^{2}\right)+x^{2} y^{2}\left(x^{4}+y^{4}\right)+$ $x^{4} y^{4}\left(x^{2}+y^{2}\right) \Leftrightarrow$
$\left(1-x^{2} y^{2}-x^{4} y^{4}+x^{6} y^{6}\right)+x^{3} y^{3}\left[x^{4}+y^{4}-x y\left(x^{2}+y^{2}\right)\right]+$ $x^{3} y^{3}\left[x^{6}+y^{6}-2 x^{3} y^{3}\right] \geqslant x^{2} y^{2}\left(x^{2}+y^{2}-2 x y\right)+$ $x^{2} y^{2}\left[x^{4}+y^{4}-x y\left(x^{2}+y^{2}\right)\right] \Leftrightarrow$ $\left(1-x^{2} y^{2}\right)^{2}\left(1+x^{2} y^{2}\right)+x^{3} y^{3}(x-y)^{2}\left(x^{2}+y^{2}+x y\right)+$ $x^{3} y^{3}(x-y)^{2}\left(x^{2}+y^{2}+x y\right)^{2} \geqslant x^{2} y^{2}(x-y)^{2}+$ $x^{2} y^{2}(x-y)^{2}\left(x^{2}+y^{2}+x y\right) \Leftrightarrow$ $\left(1-x^{2} y^{2}\right)^{2}\left(1+x^{2} y^{2}\right)+x^{3} y^{3}(x-y)^{2}\left(x^{2}+y^{2}+x y\right)\left(x^{2}+y^{2}+x y+1\right) \geqslant$ $x^{2} y^{2}(x-y)^{2}\left(x^{2}+y^{2}+x y+1\right) \Leftrightarrow$ $(1-x y)^{2}(1+x y)^{2}\left(1+x^{2} y^{2}\right) \geqslant x^{2} y^{2}(x-y)^{2}\left(x^{2}+y^{2}+x y+1\right)$ $\left[1-x y\left(x^{2}+y^{2}+x y\right)\right]$
If $1-x y\left(x^{2}+y^{2}+x y\right) \leqslant 0$, then the above inequality clearly holds; if $1-x y\left(x^{2}+y^{2}+x y\right) \geqslant$ 0, in this case
$$\begin{aligned} \text { the right side of the above inequality }= & x^{2} y^{2}(x-y)^{2}\left(x^{2}+y^{2}+x y+1\right)\left[1-x y\left(x^{2}+y^{2}+x y\right)\right]= \\ & \frac{1}{2}\left[x y\left(x^{2}+x y+y^{2}\right)-3 x^{2} y^{2}\right]\left[x y\left(x^{2}+x y+y^{2}\right)+x y\right] \\ & {\left[2-2 x y\left(x^{2}+x y+y^{2}\right)\right] \leqslant \frac{1}{2} \cdot\left(\frac{2+x y-3 x^{2} y^{2}}{3}\right)^{3}=} \end{aligned}$$
(Applying the AM-GM inequality)
$$\frac{(1-x y)^{3}(2+3 x y)^{3}}{54}$$

Therefore, it suffices to prove
$$(1-x y)(2+3 x y)^{3} \leqslant 54(1+x y)^{2}\left(1+x^{2} y^{2}\right)$$

This inequality is easily obtained from the following, i.e.,
$$\begin{aligned} 54(1+x y)^{2}\left(1+x^{2} y^{2}\right) \geqslant & 27(1+x y)^{4}=(1+x y)(3+3 x y)^{3}> \\ & (1-x y)(2+3 x y)^{3} \end{aligned}$$

Hence, equation (※) holds, and thus equation (24) holds.