Solution:
From the AM-GM inequality we have
$$a^{3} + a^{3} + 1 \geq 3a^{2} \Rightarrow 2(a^{3} + 5) \geq 3(a^{2} + 3)$$
Using the condition $ab + bc + ca = 3$, we get
$$(a^{3} + 5) \geq 3(a^{2} + ab + bc + ca) = 3(c + a)(a + b)$$
therefore
$$\frac{a}{\sqrt{a^{3} + 5}} \leq \sqrt{\frac{2a^{2}}{3(c + a)(a + b)}}$$
Using again the AM-GM inequality we get
$$\sqrt{\frac{2a^{2}}{3(c + a)(a + b)}} \leq \sqrt{\frac{2}{3}} \left( \frac{\frac{a}{c + a} + \frac{a}{a + b}}{2} \right) = \frac{\sqrt{6}}{6} \left( \frac{a}{c + a} + \frac{a}{a + b} \right)$$
From (1) and (2) we obtain
$$\frac{a}{\sqrt{a^{3} + 5}} \leq \frac{\sqrt{6}}{6} \left( \frac{a}{c + a} + \frac{a}{a + b} \right)$$
Similar inequalities hold by cyclic permutations of the $a, b, c$. Adding all these we get
$$\sum_{\text{cyclic}} \frac{a}{\sqrt{a^{3} + 5}} \leq \sum_{\text{cyc}} \frac{\sqrt{6}}{6} \left( \frac{a}{c + a} + \frac{a}{a + b} \right) = \frac{\sqrt{6}}{6} \cdot 3 = \frac{\sqrt{6}}{2}$$
which is the desired result.