$a^3 + b^3 \ge ab(a + b)$. So we have
$$\frac{\sqrt{a^3 + b^3}}{ab + 1} \ge \frac{\sqrt{ab(a + b)}}{ab + 1}$$
After substitution $a = \frac{1}{x}$, $b = \frac{1}{y}$ and $c = \frac{1}{z}$, it is sufficient to prove that for any positive real numbers $x$, $y$ and $z$ such that $xy + yz + zx = 1$, we have
$$\sum_{\text{cyc}} \frac{x}{1 + x^2} \le \frac{1}{3\sqrt{2xyz}} \sum_{\text{cyc}} \frac{\sqrt{x + y}}{1 + xy} \quad (*)$$
Since $xy + yz + zx = 1$, for the left hand side we have
$$\frac{x}{x^2 + 1} = \frac{x}{x^2 + xy + xz + yz} = \frac{x}{(x + y)(x + z)} = \frac{x(y + z)}{(x + y)(y + z)(z + x)}$$
And thus,
$$\sum_{\text{cyc}} \frac{x}{1 + x^2} = \frac{1}{(x + y)(y + z)(z + x)} \sum_{\text{cyc}} (xy + xz) = \frac{2}{(x + y)(y + z)(z + x)}$$
Therefore, (*) is equivalent to
$$\frac{2}{(x + y)(y + z)(z + x)} \le \frac{1}{3\sqrt{2xyz}} \sum_{\text{cyc}} \frac{\sqrt{x + y}}{1 + xy}$$
On the other hand, by Cauchy-Schwarz inequality, we have $(1 + x^2)(1 + y^2) \ge (1 + xy)^2$ and so
$$\begin{aligned} \frac{1}{3\sqrt{2xyz}} \sum_{\text{cyc}} \frac{\sqrt{x + y}}{1 + xy} &\ge \frac{1}{3\sqrt{2xyz}} \sum_{\text{cyc}} \frac{\sqrt{x + y}}{\sqrt{(1 + x^2)(1 + y^2)}} \\ &= \frac{1}{3\sqrt{2xyz}} \sum_{\text{cyc}} \frac{\sqrt{x + y}}{\sqrt{(x + y)^2(x + z)(y + z)}} \\ &= \frac{1}{\sqrt{2xyz(x + y)(y + z)(z + x)}} \\ &\ge \frac{2}{(x + y)(y + z)(z + x)} \end{aligned}$$
The last inequality was true because we have $(x + y)(y + z)(z + x) \ge 8xyz$, which is an immediate consequence of AM-GM inequality.