39. After factorization and applying the Cauchy-Schwarz inequality, we get
$$\begin{array}{l} \left(\frac{1}{a^{3}}-a^{2}\right)\left(\frac{1}{b^{3}}-b^{2}\right)=\frac{\left(1-a^{5}\right)\left(1-b^{5}\right)}{a^{3} b^{3}}= \\ \frac{(1-a)(1-b)\left(1+a+a^{2}+a^{3}+a^{4}\right)\left(1+b+b^{2}+b^{3}+\right.}{a^{3} b^{3}}+ \\ \frac{\left(1+a+a^{2}+a^{3}+a^{4}\right)\left(1+b+b^{2}+b^{3}+b^{4}\right)}{a^{2} b^{2}}= \\ \left(\frac{1}{a^{2}}+\frac{1}{a}+1+a+a^{2}\right)\left(\frac{1}{b^{2}}+\frac{1}{b}+1+b+b^{2}\right) \geqslant \\ \left(\frac{1}{a b}+\frac{1}{\sqrt{a b}}+1+\sqrt{a b}+a b\right)^{2}= \\ \left(a b+\frac{1}{a b}+\sqrt{a b}+\frac{1}{\sqrt{a b}}+1\right)^{2} \end{array}$$

By the AM-GM inequality, we have $\sqrt{a b} \leqslant \frac{a+b}{2}=\frac{1}{2}, a b \leqslant \frac{1}{4}$. Therefore,
$$\begin{array}{c} a b+\frac{1}{a b}-\frac{17}{4}=\frac{4 a^{2} b^{2}-17 a b+4}{4 a b}=\frac{(4 a b-1)(a b-4)}{4 a b} \geqslant 0 \\ \sqrt{a b}+\frac{1}{\sqrt{a b}}-\frac{5}{2}=\frac{2(\sqrt{a b}-1)(\sqrt{a b}-2)}{\sqrt{a b}} \geqslant 0 \end{array}$$

Thus, $a b+\frac{1}{a b}+\sqrt{a b}+\frac{1}{\sqrt{a b}}+1 \geqslant \frac{31}{4}$. The inequality is proved.